reshaping data frame and applying calculation for each row
2
I have a data frame as follows:
df=pd.DataFrame({ 'family' : ["A","A","B","B"],
'V1' : [5,5,40,10,],
'V2' :[50,10,180,20],
'gr_0' :["all","all","all","all"],
'gr_1' :["m1","m1","m2","m3"],
'gr_2' :["m12","m12","m12","m9"],
'gr_3' :["NO","m14","m15","NO"]
})
and I would like to transform it in the following way:
df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
"calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]
})
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m2 4.5
6 B m3 2.0
7 B m12 4.5
8 B m9 2.0
9 B m15 4.5
In order to reach to df_new:
- I want the rows to be aligned by "family" X each unique value of "gr_" columns.
- For every line to calculate the respective sum(V2)/sum(V1) as shown in the df_new.
I am quite new to python. Softcoding of this seems quite complex to me.
Preferably, I don't want "No" records to be listed in this df_new but it can stay in the output as well.
python python-3.x pandas pivot-table pandas-groupby
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
add a comment |
2
I have a data frame as follows:
df=pd.DataFrame({ 'family' : ["A","A","B","B"],
'V1' : [5,5,40,10,],
'V2' :[50,10,180,20],
'gr_0' :["all","all","all","all"],
'gr_1' :["m1","m1","m2","m3"],
'gr_2' :["m12","m12","m12","m9"],
'gr_3' :["NO","m14","m15","NO"]
})
and I would like to transform it in the following way:
df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
"calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]
})
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m2 4.5
6 B m3 2.0
7 B m12 4.5
8 B m9 2.0
9 B m15 4.5
In order to reach to df_new:
- I want the rows to be aligned by "family" X each unique value of "gr_" columns.
- For every line to calculate the respective sum(V2)/sum(V1) as shown in the df_new.
I am quite new to python. Softcoding of this seems quite complex to me.
Preferably, I don't want "No" records to be listed in this df_new but it can stay in the output as well.
python python-3.x pandas pivot-table pandas-groupby
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
add a comment |
2
2
2
1
I have a data frame as follows:
df=pd.DataFrame({ 'family' : ["A","A","B","B"],
'V1' : [5,5,40,10,],
'V2' :[50,10,180,20],
'gr_0' :["all","all","all","all"],
'gr_1' :["m1","m1","m2","m3"],
'gr_2' :["m12","m12","m12","m9"],
'gr_3' :["NO","m14","m15","NO"]
})
and I would like to transform it in the following way:
df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
"calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]
})
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m2 4.5
6 B m3 2.0
7 B m12 4.5
8 B m9 2.0
9 B m15 4.5
In order to reach to df_new:
- I want the rows to be aligned by "family" X each unique value of "gr_" columns.
- For every line to calculate the respective sum(V2)/sum(V1) as shown in the df_new.
I am quite new to python. Softcoding of this seems quite complex to me.
Preferably, I don't want "No" records to be listed in this df_new but it can stay in the output as well.
python python-3.x pandas pivot-table pandas-groupby
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
I have a data frame as follows:
df=pd.DataFrame({ 'family' : ["A","A","B","B"],
'V1' : [5,5,40,10,],
'V2' :[50,10,180,20],
'gr_0' :["all","all","all","all"],
'gr_1' :["m1","m1","m2","m3"],
'gr_2' :["m12","m12","m12","m9"],
'gr_3' :["NO","m14","m15","NO"]
})
and I would like to transform it in the following way:
df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
"calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]
})
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m2 4.5
6 B m3 2.0
7 B m12 4.5
8 B m9 2.0
9 B m15 4.5
In order to reach to df_new:
- I want the rows to be aligned by "family" X each unique value of "gr_" columns.
- For every line to calculate the respective sum(V2)/sum(V1) as shown in the df_new.
I am quite new to python. Softcoding of this seems quite complex to me.
Preferably, I don't want "No" records to be listed in this df_new but it can stay in the output as well.
python python-3.x pandas pivot-table pandas-groupby
python python-3.x pandas pivot-table pandas-groupby
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
edited Nov 15 '18 at 17:55
user3483203
31.4k82656
31.4k82656
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
asked Nov 15 '18 at 16:22
glsvmkyglsvmky
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
2
You can do so with this:
df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
print(df_new)
family value calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
add a comment |
1
melt + groupby:
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
A similar approach to this answer, but with the benefit of being fully vectorized, and avoiding apply
Performance:
a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
@AlexandreNixon similar in approach, but I just added timings to show whyapplyshould really be avoided. Vectorized operations make a big difference
– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
You can do so with this:
df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
print(df_new)
family value calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
add a comment |
2
You can do so with this:
df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
print(df_new)
family value calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
add a comment |
2
2
2
You can do so with this:
df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
print(df_new)
family value calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
You can do so with this:
df_new = df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)')
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
print(df_new)
family value calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
edited Nov 15 '18 at 16:47
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
answered Nov 15 '18 at 16:33
yatuyatu
13.2k31341
13.2k31341
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
add a comment |
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
1
1
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
Thanks for that @Ben.T :)
– yatu
Nov 15 '18 at 16:47
add a comment |
1
melt + groupby:
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
A similar approach to this answer, but with the benefit of being fully vectorized, and avoiding apply
Performance:
a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
@AlexandreNixon similar in approach, but I just added timings to show whyapplyshould really be avoided. Vectorized operations make a big difference
– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
add a comment |
1
melt + groupby:
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
A similar approach to this answer, but with the benefit of being fully vectorized, and avoiding apply
Performance:
a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
@AlexandreNixon similar in approach, but I just added timings to show whyapplyshould really be avoided. Vectorized operations make a big difference
– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
add a comment |
1
1
1
melt + groupby:
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
A similar approach to this answer, but with the benefit of being fully vectorized, and avoiding apply
Performance:
a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
melt + groupby:
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m12 4.5
6 B m15 4.5
7 B m2 4.5
8 B m3 2.0
9 B m9 2.0
A similar approach to this answer, but with the benefit of being fully vectorized, and avoiding apply
Performance:
a = np.random.randint(1, 1000, (1_000_000, 7))
df = pd.DataFrame(a, columns=['family', 'V1', 'V2', 'gr_0', 'gr_1', 'gr_2', 'gr_3'])
df[['gr_0', 'gr_1', 'gr_2', 'gr_3']] = df[['gr_0', 'gr_1', 'gr_2', 'gr_3']].astype(str)
%%timeit
v = df.melt(id_vars=['family','V1','V2'], value_name='gr')
w = v.loc[v.gr != 'NO']
x = w.groupby(['family', 'gr']).sum()
(x.V2 / x.V1).reset_index(name='calc(sumV2/sumV1)')
2.71 s ± 32.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
df_new = (df.melt(id_vars=['family','V1','V2']).groupby(['family','value'])
.apply(lambda x: x.V2.sum()/x.V1.sum())
.reset_index(name='calc(sumV2/sumV1)'))
df_new = df_new[df_new.value != 'NO'].reset_index(drop=True)
5min 24s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
edited Nov 15 '18 at 17:53
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
answered Nov 15 '18 at 16:47
user3483203user3483203
31.4k82656
31.4k82656
@AlexandreNixon similar in approach, but I just added timings to show whyapplyshould really be avoided. Vectorized operations make a big difference
– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
add a comment |
@AlexandreNixon similar in approach, but I just added timings to show whyapplyshould really be avoided. Vectorized operations make a big difference
– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
@AlexandreNixon similar in approach, but I just added timings to show why
apply should really be avoided. Vectorized operations make a big difference– user3483203
Nov 15 '18 at 17:54
@AlexandreNixon similar in approach, but I just added timings to show why
apply should really be avoided. Vectorized operations make a big difference– user3483203
Nov 15 '18 at 17:54
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
Fair point @user3483203
– yatu
Nov 15 '18 at 19:01
add a comment |
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