Create FY based on the range of date in pandas
I am working on dates and FY.
My working data set contains ten thousands of rows with date.
I want to create a new column to identify the season or FY.
The sample data frame is as follow:
df = pd.DataFrame()
df['date'] = ['10/08/2018','12/09/2018','15/08/2017','16/05/2018']
what I wanted to do is to create a new column season based on the range of the date.
for example,
if the month of the date is from AUgust to June, it will be considered as a season.
It means that the date ranges from 01/08/2005 to 30/06/2006, the season will be 2005-06.
For the sample date frame, the expected output will be as follow:
date season
10/082018 2018-19
12/09/2018 2018-19
15/08/2017 2017-18
16/05/2018 2017-18
Below is my expected output:
How would I define the range and the season??
Thanks,
Zep.
python pandas date date-range
asked Nov 13 '18 at 4:58
Zephyr
42810
add a comment |
I am working on dates and FY.
My working data set contains ten thousands of rows with date.
I want to create a new column to identify the season or FY.
The sample data frame is as follow:
df = pd.DataFrame()
df['date'] = ['10/08/2018','12/09/2018','15/08/2017','16/05/2018']
what I wanted to do is to create a new column season based on the range of the date.
for example,
if the month of the date is from AUgust to June, it will be considered as a season.
It means that the date ranges from 01/08/2005 to 30/06/2006, the season will be 2005-06.
For the sample date frame, the expected output will be as follow:
date season
10/082018 2018-19
12/09/2018 2018-19
15/08/2017 2017-18
16/05/2018 2017-18
Below is my expected output:
How would I define the range and the season??
Thanks,
Zep.
python pandas date date-range
asked Nov 13 '18 at 4:58
Zephyr
42810
Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02
add a comment |
I am working on dates and FY.
My working data set contains ten thousands of rows with date.
I want to create a new column to identify the season or FY.
The sample data frame is as follow:
df = pd.DataFrame()
df['date'] = ['10/08/2018','12/09/2018','15/08/2017','16/05/2018']
what I wanted to do is to create a new column season based on the range of the date.
for example,
if the month of the date is from AUgust to June, it will be considered as a season.
It means that the date ranges from 01/08/2005 to 30/06/2006, the season will be 2005-06.
For the sample date frame, the expected output will be as follow:
date season
10/082018 2018-19
12/09/2018 2018-19
15/08/2017 2017-18
16/05/2018 2017-18
Below is my expected output:
How would I define the range and the season??
Thanks,
Zep.
python pandas date date-range
asked Nov 13 '18 at 4:58
Zephyr
42810
I am working on dates and FY.
My working data set contains ten thousands of rows with date.
I want to create a new column to identify the season or FY.
The sample data frame is as follow:
df = pd.DataFrame()
df['date'] = ['10/08/2018','12/09/2018','15/08/2017','16/05/2018']
what I wanted to do is to create a new column season based on the range of the date.
for example,
if the month of the date is from AUgust to June, it will be considered as a season.
It means that the date ranges from 01/08/2005 to 30/06/2006, the season will be 2005-06.
For the sample date frame, the expected output will be as follow:
date season
10/082018 2018-19
12/09/2018 2018-19
15/08/2017 2017-18
16/05/2018 2017-18
Below is my expected output:
How would I define the range and the season??
Thanks,
Zep.
python pandas date date-range
python pandas date date-range
asked Nov 13 '18 at 4:58
Zephyr
42810
asked Nov 13 '18 at 4:58
Zephyr
42810
edited Nov 13 '18 at 5:05
asked Nov 13 '18 at 4:58
Zephyr
42810
asked Nov 13 '18 at 4:58
Zephyr
42810
asked Nov 13 '18 at 4:58
Zephyr
42810
42810
Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02
add a comment |
Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02
Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02
Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02
add a comment |
1 Answer
1
active
oldest
votes
You can use the condition based on month and concatenate the year part of the date after converting to string
df.date = pd.to_datetime(df.date, format = '%d/%m/%Y')
cond = df.date.dt.month >=8
df['season'] = np.where(cond, df.date.dt.year.apply(str) + '-' + (df.date.dt.year+1).apply(str).str[2:], (df.date.dt.year-1).apply(str) + '-' + df.date.dt.year.apply(str).str[2:])
date season
0 2018-08-10 2018-19
1 2018-09-12 2018-19
2 2017-08-15 2017-18
3 2018-05-16 2017-18
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
@Vaishali I would suggest to useyear = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:])to speed up the process since you have lot of intermediate same variables.
– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
add a comment |
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1 Answer
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You can use the condition based on month and concatenate the year part of the date after converting to string
df.date = pd.to_datetime(df.date, format = '%d/%m/%Y')
cond = df.date.dt.month >=8
df['season'] = np.where(cond, df.date.dt.year.apply(str) + '-' + (df.date.dt.year+1).apply(str).str[2:], (df.date.dt.year-1).apply(str) + '-' + df.date.dt.year.apply(str).str[2:])
date season
0 2018-08-10 2018-19
1 2018-09-12 2018-19
2 2017-08-15 2017-18
3 2018-05-16 2017-18
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
@Vaishali I would suggest to useyear = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:])to speed up the process since you have lot of intermediate same variables.
– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
add a comment |
You can use the condition based on month and concatenate the year part of the date after converting to string
df.date = pd.to_datetime(df.date, format = '%d/%m/%Y')
cond = df.date.dt.month >=8
df['season'] = np.where(cond, df.date.dt.year.apply(str) + '-' + (df.date.dt.year+1).apply(str).str[2:], (df.date.dt.year-1).apply(str) + '-' + df.date.dt.year.apply(str).str[2:])
date season
0 2018-08-10 2018-19
1 2018-09-12 2018-19
2 2017-08-15 2017-18
3 2018-05-16 2017-18
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
@Vaishali I would suggest to useyear = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:])to speed up the process since you have lot of intermediate same variables.
– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
add a comment |
You can use the condition based on month and concatenate the year part of the date after converting to string
df.date = pd.to_datetime(df.date, format = '%d/%m/%Y')
cond = df.date.dt.month >=8
df['season'] = np.where(cond, df.date.dt.year.apply(str) + '-' + (df.date.dt.year+1).apply(str).str[2:], (df.date.dt.year-1).apply(str) + '-' + df.date.dt.year.apply(str).str[2:])
date season
0 2018-08-10 2018-19
1 2018-09-12 2018-19
2 2017-08-15 2017-18
3 2018-05-16 2017-18
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
You can use the condition based on month and concatenate the year part of the date after converting to string
df.date = pd.to_datetime(df.date, format = '%d/%m/%Y')
cond = df.date.dt.month >=8
df['season'] = np.where(cond, df.date.dt.year.apply(str) + '-' + (df.date.dt.year+1).apply(str).str[2:], (df.date.dt.year-1).apply(str) + '-' + df.date.dt.year.apply(str).str[2:])
date season
0 2018-08-10 2018-19
1 2018-09-12 2018-19
2 2017-08-15 2017-18
3 2018-05-16 2017-18
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
answered Nov 13 '18 at 5:21
Vaishali
17.8k31028
17.8k31028
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
@Vaishali I would suggest to useyear = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:])to speed up the process since you have lot of intermediate same variables.
– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
add a comment |
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
@Vaishali I would suggest to useyear = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:])to speed up the process since you have lot of intermediate same variables.
– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
Thanks Vaishali. I converted the Date to datetime and got with the following format. time data Timestamp('2018-08-10 00:00:00') how can I remove the time part from date time? Thanks for your help.
– Zephyr
Nov 13 '18 at 5:29
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
@Zephyr, df['date'].dt.date will give you the date component of datetime
– Vaishali
Nov 13 '18 at 5:30
1
1
@Vaishali I would suggest to use
year = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:]) to speed up the process since you have lot of intermediate same variables.– Sandeep Kadapa
Nov 13 '18 at 5:37
@Vaishali I would suggest to use
year = df.date.dt.year;year_str = df.date.dt.year.astype(str);np.where(cond,year_str + '-' + (year+1).astype(str).str[2:], (year-1).astype(str) + '-' + year_str.str[2:]) to speed up the process since you have lot of intermediate same variables.– Sandeep Kadapa
Nov 13 '18 at 5:37
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
@SandeepKadapa, may be you compared the time for the given number of rows. Try it on df = pd.concat([df]*10000, ignore_index=True) and times are identical.
– Vaishali
Nov 13 '18 at 5:44
1
1
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
@Vaishali Oh that's interesting and thank you for letting me know that.
– Sandeep Kadapa
Nov 13 '18 at 5:55
add a comment |
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Please provide a Minimal, Complete, and Verifiable example. Images are difficult to interpret.
– Sandeep Kadapa
Nov 13 '18 at 5:02