Writing one tuple to a CSV in multiple rows












0















So, I have a python tuple. I am trying to write it to a CSV file. What I did so far I am able to write to the CSV but all my data appears in one row. Can anyone tell me how can I convert it into columns ? For example this is the data in one row on my CSV.



100 0.01 11139155 5569534 0.499996 NO SYNC 555 0.01 2306110 1153050 0.499998 NO SYNC 333 0.22 3434535 4446466 0.453535 NO SYNC


What I want to do I want to organize this tuple in a way that after each NO SYNC it moves to the next row.



100 0.01 11139155 5569534 0.499996 NO SYNC
555 0.01 2306110 1153050 0.499998 NO SYNC
333 0.22 3434535 4446466 0.453535 NO SYNC


This is my script



with open ('File path') as f:
writer = csv.writer(f, delimiter = ',')
writer.writerow(results_t1)
f.close()


Where results_t1 is my input tuple.



My input tuple looks like this :



(100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')










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  • 1





    What do your input tuples look like? what code have you tried so far?

    – Alex
    Nov 14 '18 at 15:57






  • 1





    And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

    – Jon Kiparsky
    Nov 14 '18 at 15:59











  • Yes, this is what I want to write to a file with a new row after NO SYNC

    – Omar
    Nov 14 '18 at 16:01











  • @Omar - any luck with these implementations?

    – rs311
    Nov 15 '18 at 0:34
















0















So, I have a python tuple. I am trying to write it to a CSV file. What I did so far I am able to write to the CSV but all my data appears in one row. Can anyone tell me how can I convert it into columns ? For example this is the data in one row on my CSV.



100 0.01 11139155 5569534 0.499996 NO SYNC 555 0.01 2306110 1153050 0.499998 NO SYNC 333 0.22 3434535 4446466 0.453535 NO SYNC


What I want to do I want to organize this tuple in a way that after each NO SYNC it moves to the next row.



100 0.01 11139155 5569534 0.499996 NO SYNC
555 0.01 2306110 1153050 0.499998 NO SYNC
333 0.22 3434535 4446466 0.453535 NO SYNC


This is my script



with open ('File path') as f:
writer = csv.writer(f, delimiter = ',')
writer.writerow(results_t1)
f.close()


Where results_t1 is my input tuple.



My input tuple looks like this :



(100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')










share|improve this question




















  • 1





    What do your input tuples look like? what code have you tried so far?

    – Alex
    Nov 14 '18 at 15:57






  • 1





    And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

    – Jon Kiparsky
    Nov 14 '18 at 15:59











  • Yes, this is what I want to write to a file with a new row after NO SYNC

    – Omar
    Nov 14 '18 at 16:01











  • @Omar - any luck with these implementations?

    – rs311
    Nov 15 '18 at 0:34














0












0








0








So, I have a python tuple. I am trying to write it to a CSV file. What I did so far I am able to write to the CSV but all my data appears in one row. Can anyone tell me how can I convert it into columns ? For example this is the data in one row on my CSV.



100 0.01 11139155 5569534 0.499996 NO SYNC 555 0.01 2306110 1153050 0.499998 NO SYNC 333 0.22 3434535 4446466 0.453535 NO SYNC


What I want to do I want to organize this tuple in a way that after each NO SYNC it moves to the next row.



100 0.01 11139155 5569534 0.499996 NO SYNC
555 0.01 2306110 1153050 0.499998 NO SYNC
333 0.22 3434535 4446466 0.453535 NO SYNC


This is my script



with open ('File path') as f:
writer = csv.writer(f, delimiter = ',')
writer.writerow(results_t1)
f.close()


Where results_t1 is my input tuple.



My input tuple looks like this :



(100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')










share|improve this question
















So, I have a python tuple. I am trying to write it to a CSV file. What I did so far I am able to write to the CSV but all my data appears in one row. Can anyone tell me how can I convert it into columns ? For example this is the data in one row on my CSV.



100 0.01 11139155 5569534 0.499996 NO SYNC 555 0.01 2306110 1153050 0.499998 NO SYNC 333 0.22 3434535 4446466 0.453535 NO SYNC


What I want to do I want to organize this tuple in a way that after each NO SYNC it moves to the next row.



100 0.01 11139155 5569534 0.499996 NO SYNC
555 0.01 2306110 1153050 0.499998 NO SYNC
333 0.22 3434535 4446466 0.453535 NO SYNC


This is my script



with open ('File path') as f:
writer = csv.writer(f, delimiter = ',')
writer.writerow(results_t1)
f.close()


Where results_t1 is my input tuple.



My input tuple looks like this :



(100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')







python






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 '18 at 11:00









Rahul Chawla

563512




563512










asked Nov 14 '18 at 15:55









OmarOmar

62




62








  • 1





    What do your input tuples look like? what code have you tried so far?

    – Alex
    Nov 14 '18 at 15:57






  • 1





    And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

    – Jon Kiparsky
    Nov 14 '18 at 15:59











  • Yes, this is what I want to write to a file with a new row after NO SYNC

    – Omar
    Nov 14 '18 at 16:01











  • @Omar - any luck with these implementations?

    – rs311
    Nov 15 '18 at 0:34














  • 1





    What do your input tuples look like? what code have you tried so far?

    – Alex
    Nov 14 '18 at 15:57






  • 1





    And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

    – Jon Kiparsky
    Nov 14 '18 at 15:59











  • Yes, this is what I want to write to a file with a new row after NO SYNC

    – Omar
    Nov 14 '18 at 16:01











  • @Omar - any luck with these implementations?

    – rs311
    Nov 15 '18 at 0:34








1




1





What do your input tuples look like? what code have you tried so far?

– Alex
Nov 14 '18 at 15:57





What do your input tuples look like? what code have you tried so far?

– Alex
Nov 14 '18 at 15:57




1




1





And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

– Jon Kiparsky
Nov 14 '18 at 15:59





And you want it to be space-delimited rather than comma-delimited? (ie, is that the actual output you're looking to write to a file?)

– Jon Kiparsky
Nov 14 '18 at 15:59













Yes, this is what I want to write to a file with a new row after NO SYNC

– Omar
Nov 14 '18 at 16:01





Yes, this is what I want to write to a file with a new row after NO SYNC

– Omar
Nov 14 '18 at 16:01













@Omar - any luck with these implementations?

– rs311
Nov 15 '18 at 0:34





@Omar - any luck with these implementations?

– rs311
Nov 15 '18 at 0:34












3 Answers
3






active

oldest

votes


















1














Assuming constant number of columns in CSV (Which usually is the case). And using chunk breaker got from a highly coveted answer here



d = (100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')

# 6 columns in a row
chunks = [d[i:i + 6] for i in range(0, len(d), 6)]

with open ('File path', 'w') as f:
writer = csv.writer(f, delimiter = ',')
writer.writerows(d)





share|improve this answer


























  • i think you need to add with open ('File path', 'w) as f:

    – rs311
    Nov 14 '18 at 16:41











  • I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

    – Omar
    Nov 14 '18 at 16:51











  • @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

    – Rahul Chawla
    Nov 16 '18 at 8:17



















0














Here's a quick example of one way you might approach this.



>>> data = "1 2 3 4 5 6 7 8 9 10"
>>> items = data.split(" ")
>>> chunk_size = 5
>>> chunks = [items[i:i+chunk_size] for i in range(len(items)/chunk_size)]
>>> chunks
[['1', '2', '3', '4', '5'], ['2', '3', '4', '5', '6']]
>>> "n".join([" ".join(chunk) for chunk in chunks])
'1 2 3 4 5n2 3 4 5 6'


Note that there are some unsafe assumptions here. For example, I'm assuming that your lines are of a consistent length which you know at the start.



If your lines are not of a known consistent length, and you want to break on a particular token, here's how you might do that:



>>> data = "1 2 3 4 STOP 5 6 7 8 STOP"
>>> token = "STOP"
>>> chunks = data.split(token)
>>> "n".join(["{} {}".format(chunk, token) for chunk in chunks])
'1 2 3 4 STOPn 5 6 7 8 STOPn STOP'
>>>


Obviously, once you have the data rows you can simply emit them to a file.






share|improve this answer































    0














    I may have gone a little overboard on the solution here, but the below will provide you the output you're looking to achieve.



    This should provide a re-usable function for the future in case you have another problem similar to this which forces you to find something other than 'NO SYNC'.



    Like @Rahul Chawla's answer, this looks for your new line identifier NO SYNC and updates it to be NOSYNC!. The reason this is helpful is that it gives us a way to create a list of entries ending with NO SYNC by splitting a string every time we see !. You can customize what delimiter (delim paramter, default is !) gets appended to your find parameter based on what is not present in your data.



    import csv

    raw_data = (
    100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC',
    3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC',
    94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169,
    'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304,
    0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040,
    710518, 0.499998592580082, 'NO SYNC')


    def identify_new_rows(data, find, delim='!'):
    """convert a flat tuple/list, or string into a list of lists given something to `find`."""
    if isinstance(data, tuple):
    data = " ".join(map(str, data))
    elif isinstance(data, list):
    data = " ".join(data)
    data = [[x for x in d.split(' ') if str(x) != ''] for d in data.replace(find, str(find).replace(' ', '') + delim).split(delim) if len(d) > 0]
    return data



    with open(r'~/path/to/your/file.csv', 'w', newline='') as myfile:
    wr = csv.writer(myfile, quoting=csv.QUOTE_ALL, delimiter=',')
    wr.writerows(identify_new_rows(raw_data, 'NO SYNC'))


    I will note there are flaws in this approach when attempting to apply to other data sets, as this presumes there are no strings with spaces in your data (other than the current delimiter - which gets handled in the list comprehension.






    share|improve this answer

























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      3 Answers
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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      1














      Assuming constant number of columns in CSV (Which usually is the case). And using chunk breaker got from a highly coveted answer here



      d = (100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')

      # 6 columns in a row
      chunks = [d[i:i + 6] for i in range(0, len(d), 6)]

      with open ('File path', 'w') as f:
      writer = csv.writer(f, delimiter = ',')
      writer.writerows(d)





      share|improve this answer


























      • i think you need to add with open ('File path', 'w) as f:

        – rs311
        Nov 14 '18 at 16:41











      • I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

        – Omar
        Nov 14 '18 at 16:51











      • @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

        – Rahul Chawla
        Nov 16 '18 at 8:17
















      1














      Assuming constant number of columns in CSV (Which usually is the case). And using chunk breaker got from a highly coveted answer here



      d = (100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')

      # 6 columns in a row
      chunks = [d[i:i + 6] for i in range(0, len(d), 6)]

      with open ('File path', 'w') as f:
      writer = csv.writer(f, delimiter = ',')
      writer.writerows(d)





      share|improve this answer


























      • i think you need to add with open ('File path', 'w) as f:

        – rs311
        Nov 14 '18 at 16:41











      • I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

        – Omar
        Nov 14 '18 at 16:51











      • @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

        – Rahul Chawla
        Nov 16 '18 at 8:17














      1












      1








      1







      Assuming constant number of columns in CSV (Which usually is the case). And using chunk breaker got from a highly coveted answer here



      d = (100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')

      # 6 columns in a row
      chunks = [d[i:i + 6] for i in range(0, len(d), 6)]

      with open ('File path', 'w') as f:
      writer = csv.writer(f, delimiter = ',')
      writer.writerows(d)





      share|improve this answer















      Assuming constant number of columns in CSV (Which usually is the case). And using chunk breaker got from a highly coveted answer here



      d = (100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC', 3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC', 94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169, 'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304, 0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040, 710518, 0.499998592580082, 'NO SYNC')

      # 6 columns in a row
      chunks = [d[i:i + 6] for i in range(0, len(d), 6)]

      with open ('File path', 'w') as f:
      writer = csv.writer(f, delimiter = ',')
      writer.writerows(d)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 16 '18 at 11:12

























      answered Nov 14 '18 at 16:12









      Rahul ChawlaRahul Chawla

      563512




      563512













      • i think you need to add with open ('File path', 'w) as f:

        – rs311
        Nov 14 '18 at 16:41











      • I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

        – Omar
        Nov 14 '18 at 16:51











      • @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

        – Rahul Chawla
        Nov 16 '18 at 8:17



















      • i think you need to add with open ('File path', 'w) as f:

        – rs311
        Nov 14 '18 at 16:41











      • I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

        – Omar
        Nov 14 '18 at 16:51











      • @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

        – Rahul Chawla
        Nov 16 '18 at 8:17

















      i think you need to add with open ('File path', 'w) as f:

      – rs311
      Nov 14 '18 at 16:41





      i think you need to add with open ('File path', 'w) as f:

      – rs311
      Nov 14 '18 at 16:41













      I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

      – Omar
      Nov 14 '18 at 16:51





      I am trying your technique but it is giving me error. So, NO SYNC is a string as you can see in my updated post. It is giving me an error 'tuple' object has no attribute 'replace'. Also, can you please tell me why youre using a $ sign ?

      – Omar
      Nov 14 '18 at 16:51













      @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

      – Rahul Chawla
      Nov 16 '18 at 8:17





      @rs311 Thanks for pointing that out, edited the answer now. Omar, added the explanation in answer itself.

      – Rahul Chawla
      Nov 16 '18 at 8:17













      0














      Here's a quick example of one way you might approach this.



      >>> data = "1 2 3 4 5 6 7 8 9 10"
      >>> items = data.split(" ")
      >>> chunk_size = 5
      >>> chunks = [items[i:i+chunk_size] for i in range(len(items)/chunk_size)]
      >>> chunks
      [['1', '2', '3', '4', '5'], ['2', '3', '4', '5', '6']]
      >>> "n".join([" ".join(chunk) for chunk in chunks])
      '1 2 3 4 5n2 3 4 5 6'


      Note that there are some unsafe assumptions here. For example, I'm assuming that your lines are of a consistent length which you know at the start.



      If your lines are not of a known consistent length, and you want to break on a particular token, here's how you might do that:



      >>> data = "1 2 3 4 STOP 5 6 7 8 STOP"
      >>> token = "STOP"
      >>> chunks = data.split(token)
      >>> "n".join(["{} {}".format(chunk, token) for chunk in chunks])
      '1 2 3 4 STOPn 5 6 7 8 STOPn STOP'
      >>>


      Obviously, once you have the data rows you can simply emit them to a file.






      share|improve this answer




























        0














        Here's a quick example of one way you might approach this.



        >>> data = "1 2 3 4 5 6 7 8 9 10"
        >>> items = data.split(" ")
        >>> chunk_size = 5
        >>> chunks = [items[i:i+chunk_size] for i in range(len(items)/chunk_size)]
        >>> chunks
        [['1', '2', '3', '4', '5'], ['2', '3', '4', '5', '6']]
        >>> "n".join([" ".join(chunk) for chunk in chunks])
        '1 2 3 4 5n2 3 4 5 6'


        Note that there are some unsafe assumptions here. For example, I'm assuming that your lines are of a consistent length which you know at the start.



        If your lines are not of a known consistent length, and you want to break on a particular token, here's how you might do that:



        >>> data = "1 2 3 4 STOP 5 6 7 8 STOP"
        >>> token = "STOP"
        >>> chunks = data.split(token)
        >>> "n".join(["{} {}".format(chunk, token) for chunk in chunks])
        '1 2 3 4 STOPn 5 6 7 8 STOPn STOP'
        >>>


        Obviously, once you have the data rows you can simply emit them to a file.






        share|improve this answer


























          0












          0








          0







          Here's a quick example of one way you might approach this.



          >>> data = "1 2 3 4 5 6 7 8 9 10"
          >>> items = data.split(" ")
          >>> chunk_size = 5
          >>> chunks = [items[i:i+chunk_size] for i in range(len(items)/chunk_size)]
          >>> chunks
          [['1', '2', '3', '4', '5'], ['2', '3', '4', '5', '6']]
          >>> "n".join([" ".join(chunk) for chunk in chunks])
          '1 2 3 4 5n2 3 4 5 6'


          Note that there are some unsafe assumptions here. For example, I'm assuming that your lines are of a consistent length which you know at the start.



          If your lines are not of a known consistent length, and you want to break on a particular token, here's how you might do that:



          >>> data = "1 2 3 4 STOP 5 6 7 8 STOP"
          >>> token = "STOP"
          >>> chunks = data.split(token)
          >>> "n".join(["{} {}".format(chunk, token) for chunk in chunks])
          '1 2 3 4 STOPn 5 6 7 8 STOPn STOP'
          >>>


          Obviously, once you have the data rows you can simply emit them to a file.






          share|improve this answer













          Here's a quick example of one way you might approach this.



          >>> data = "1 2 3 4 5 6 7 8 9 10"
          >>> items = data.split(" ")
          >>> chunk_size = 5
          >>> chunks = [items[i:i+chunk_size] for i in range(len(items)/chunk_size)]
          >>> chunks
          [['1', '2', '3', '4', '5'], ['2', '3', '4', '5', '6']]
          >>> "n".join([" ".join(chunk) for chunk in chunks])
          '1 2 3 4 5n2 3 4 5 6'


          Note that there are some unsafe assumptions here. For example, I'm assuming that your lines are of a consistent length which you know at the start.



          If your lines are not of a known consistent length, and you want to break on a particular token, here's how you might do that:



          >>> data = "1 2 3 4 STOP 5 6 7 8 STOP"
          >>> token = "STOP"
          >>> chunks = data.split(token)
          >>> "n".join(["{} {}".format(chunk, token) for chunk in chunks])
          '1 2 3 4 STOPn 5 6 7 8 STOPn STOP'
          >>>


          Obviously, once you have the data rows you can simply emit them to a file.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 14 '18 at 16:13









          Jon KiparskyJon Kiparsky

          5,42021633




          5,42021633























              0














              I may have gone a little overboard on the solution here, but the below will provide you the output you're looking to achieve.



              This should provide a re-usable function for the future in case you have another problem similar to this which forces you to find something other than 'NO SYNC'.



              Like @Rahul Chawla's answer, this looks for your new line identifier NO SYNC and updates it to be NOSYNC!. The reason this is helpful is that it gives us a way to create a list of entries ending with NO SYNC by splitting a string every time we see !. You can customize what delimiter (delim paramter, default is !) gets appended to your find parameter based on what is not present in your data.



              import csv

              raw_data = (
              100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC',
              3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC',
              94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169,
              'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304,
              0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040,
              710518, 0.499998592580082, 'NO SYNC')


              def identify_new_rows(data, find, delim='!'):
              """convert a flat tuple/list, or string into a list of lists given something to `find`."""
              if isinstance(data, tuple):
              data = " ".join(map(str, data))
              elif isinstance(data, list):
              data = " ".join(data)
              data = [[x for x in d.split(' ') if str(x) != ''] for d in data.replace(find, str(find).replace(' ', '') + delim).split(delim) if len(d) > 0]
              return data



              with open(r'~/path/to/your/file.csv', 'w', newline='') as myfile:
              wr = csv.writer(myfile, quoting=csv.QUOTE_ALL, delimiter=',')
              wr.writerows(identify_new_rows(raw_data, 'NO SYNC'))


              I will note there are flaws in this approach when attempting to apply to other data sets, as this presumes there are no strings with spaces in your data (other than the current delimiter - which gets handled in the list comprehension.






              share|improve this answer






























                0














                I may have gone a little overboard on the solution here, but the below will provide you the output you're looking to achieve.



                This should provide a re-usable function for the future in case you have another problem similar to this which forces you to find something other than 'NO SYNC'.



                Like @Rahul Chawla's answer, this looks for your new line identifier NO SYNC and updates it to be NOSYNC!. The reason this is helpful is that it gives us a way to create a list of entries ending with NO SYNC by splitting a string every time we see !. You can customize what delimiter (delim paramter, default is !) gets appended to your find parameter based on what is not present in your data.



                import csv

                raw_data = (
                100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC',
                3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC',
                94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169,
                'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304,
                0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040,
                710518, 0.499998592580082, 'NO SYNC')


                def identify_new_rows(data, find, delim='!'):
                """convert a flat tuple/list, or string into a list of lists given something to `find`."""
                if isinstance(data, tuple):
                data = " ".join(map(str, data))
                elif isinstance(data, list):
                data = " ".join(data)
                data = [[x for x in d.split(' ') if str(x) != ''] for d in data.replace(find, str(find).replace(' ', '') + delim).split(delim) if len(d) > 0]
                return data



                with open(r'~/path/to/your/file.csv', 'w', newline='') as myfile:
                wr = csv.writer(myfile, quoting=csv.QUOTE_ALL, delimiter=',')
                wr.writerows(identify_new_rows(raw_data, 'NO SYNC'))


                I will note there are flaws in this approach when attempting to apply to other data sets, as this presumes there are no strings with spaces in your data (other than the current delimiter - which gets handled in the list comprehension.






                share|improve this answer




























                  0












                  0








                  0







                  I may have gone a little overboard on the solution here, but the below will provide you the output you're looking to achieve.



                  This should provide a re-usable function for the future in case you have another problem similar to this which forces you to find something other than 'NO SYNC'.



                  Like @Rahul Chawla's answer, this looks for your new line identifier NO SYNC and updates it to be NOSYNC!. The reason this is helpful is that it gives us a way to create a list of entries ending with NO SYNC by splitting a string every time we see !. You can customize what delimiter (delim paramter, default is !) gets appended to your find parameter based on what is not present in your data.



                  import csv

                  raw_data = (
                  100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC',
                  3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC',
                  94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169,
                  'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304,
                  0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040,
                  710518, 0.499998592580082, 'NO SYNC')


                  def identify_new_rows(data, find, delim='!'):
                  """convert a flat tuple/list, or string into a list of lists given something to `find`."""
                  if isinstance(data, tuple):
                  data = " ".join(map(str, data))
                  elif isinstance(data, list):
                  data = " ".join(data)
                  data = [[x for x in d.split(' ') if str(x) != ''] for d in data.replace(find, str(find).replace(' ', '') + delim).split(delim) if len(d) > 0]
                  return data



                  with open(r'~/path/to/your/file.csv', 'w', newline='') as myfile:
                  wr = csv.writer(myfile, quoting=csv.QUOTE_ALL, delimiter=',')
                  wr.writerows(identify_new_rows(raw_data, 'NO SYNC'))


                  I will note there are flaws in this approach when attempting to apply to other data sets, as this presumes there are no strings with spaces in your data (other than the current delimiter - which gets handled in the list comprehension.






                  share|improve this answer















                  I may have gone a little overboard on the solution here, but the below will provide you the output you're looking to achieve.



                  This should provide a re-usable function for the future in case you have another problem similar to this which forces you to find something other than 'NO SYNC'.



                  Like @Rahul Chawla's answer, this looks for your new line identifier NO SYNC and updates it to be NOSYNC!. The reason this is helpful is that it gives us a way to create a list of entries ending with NO SYNC by splitting a string every time we see !. You can customize what delimiter (delim paramter, default is !) gets appended to your find parameter based on what is not present in your data.



                  import csv

                  raw_data = (
                  100, 0.01, 11139155, 5569534, 0.499996094856387, 'NO SYNC', 555, 0.01, 2306110, 1153050, 0.499997831846703, 'NO SYNC',
                  3081, 0.01, 1951735, 975863, 0.499997694359122, 'NO SYNC', 17100, 0.01, 2896740, 1448360, 0.499996547843438, 'NO SYNC',
                  94912, 0.01, 1800105, 900045, 0.499995833576375, 'NO SYNC', 526805, 0.01, 2290245, 1145113, 0.499995851972169,
                  'NO SYNC', 2924018, 0.01, 2256745, 1128371, 0.499999335325879, 'NO SYNC', 16229682, 0.01, 2004625, 1002304,
                  0.49999575980545, 'NO SYNC', 90082412, 0.01, 1912945, 956468, 0.499997647606178, 'NO SYNC', 500000000, 0.01, 1421040,
                  710518, 0.499998592580082, 'NO SYNC')


                  def identify_new_rows(data, find, delim='!'):
                  """convert a flat tuple/list, or string into a list of lists given something to `find`."""
                  if isinstance(data, tuple):
                  data = " ".join(map(str, data))
                  elif isinstance(data, list):
                  data = " ".join(data)
                  data = [[x for x in d.split(' ') if str(x) != ''] for d in data.replace(find, str(find).replace(' ', '') + delim).split(delim) if len(d) > 0]
                  return data



                  with open(r'~/path/to/your/file.csv', 'w', newline='') as myfile:
                  wr = csv.writer(myfile, quoting=csv.QUOTE_ALL, delimiter=',')
                  wr.writerows(identify_new_rows(raw_data, 'NO SYNC'))


                  I will note there are flaws in this approach when attempting to apply to other data sets, as this presumes there are no strings with spaces in your data (other than the current delimiter - which gets handled in the list comprehension.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 15 '18 at 12:30

























                  answered Nov 14 '18 at 19:49









                  rs311rs311

                  15310




                  15310






























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