Pandas dataframe: Remove secondary upcoming same value
9
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
48.9k116390
asked Dec 6 at 15:33
s900n
435616
add a comment |
9
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
48.9k116390
asked Dec 6 at 15:33
s900n
435616
add a comment |
9
9
9
1
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
48.9k116390
asked Dec 6 at 15:33
s900n
435616
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
python pandas dataframe
edited Dec 6 at 15:46
timgeb
48.9k116390
asked Dec 6 at 15:33
s900n
435616
edited Dec 6 at 15:46
timgeb
48.9k116390
asked Dec 6 at 15:33
s900n
435616
edited Dec 6 at 15:46
timgeb
48.9k116390
edited Dec 6 at 15:46
timgeb
48.9k116390
edited Dec 6 at 15:46
timgeb
48.9k116390
48.9k116390
asked Dec 6 at 15:33
s900n
435616
asked Dec 6 at 15:33
s900n
435616
asked Dec 6 at 15:33
s900n
435616
435616
add a comment |
add a comment |
8 Answers
8
active
oldest
votes
9
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
90.8k2052101
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
4
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
152k22144285
add a comment |
4
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
48.9k116390
add a comment |
3
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
100k73163
add a comment |
3
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
30.2k82354
add a comment |
1
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
84148
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
1
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
563420
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
0
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
90.8k2052101
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
9
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
90.8k2052101
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
9
9
9
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
90.8k2052101
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
90.8k2052101
answered Dec 6 at 15:37
jpp
90.8k2052101
answered Dec 6 at 15:37
jpp
90.8k2052101
answered Dec 6 at 15:37
jpp
90.8k2052101
90.8k2052101
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
Can you think of a good solution for the case when the index is arbitrary, like
Index(['u', 'v', 'w', 'x', 'y', 'z'] AND col2 could be something like [2, 0, 0, 1, 3, 1]?– timgeb
Dec 6 at 16:18
Can you think of a good solution for the case when the index is arbitrary, like
Index(['u', 'v', 'w', 'x', 'y', 'z'] AND col2 could be something like [2, 0, 0, 1, 3, 1]?– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something like
df.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.– jpp
Dec 6 at 16:28
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something like
df.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
4
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
152k22144285
add a comment |
4
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
152k22144285
add a comment |
4
4
4
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
152k22144285
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
152k22144285
edited Dec 6 at 15:56
answered Dec 6 at 15:51
piRSquared
152k22144285
answered Dec 6 at 15:51
piRSquared
152k22144285
answered Dec 6 at 15:51
piRSquared
152k22144285
152k22144285
add a comment |
add a comment |
4
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
48.9k116390
add a comment |
4
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
48.9k116390
add a comment |
4
4
4
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
48.9k116390
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
48.9k116390
edited Dec 6 at 16:29
answered Dec 6 at 15:38
timgeb
48.9k116390
answered Dec 6 at 15:38
timgeb
48.9k116390
answered Dec 6 at 15:38
timgeb
48.9k116390
48.9k116390
add a comment |
add a comment |
3
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
100k73163
add a comment |
3
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
100k73163
add a comment |
3
3
3
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
100k73163
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
100k73163
answered Dec 6 at 15:41
W-B
100k73163
answered Dec 6 at 15:41
W-B
100k73163
answered Dec 6 at 15:41
W-B
100k73163
100k73163
add a comment |
add a comment |
3
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
30.2k82354
add a comment |
3
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
30.2k82354
add a comment |
3
3
3
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
30.2k82354
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
30.2k82354
edited Dec 6 at 15:53
answered Dec 6 at 15:39
user3483203
30.2k82354
answered Dec 6 at 15:39
user3483203
30.2k82354
answered Dec 6 at 15:39
user3483203
30.2k82354
30.2k82354
add a comment |
add a comment |
1
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
84148
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
1
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
84148
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
1
1
1
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
84148
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
84148
answered Dec 6 at 15:42
iamklaus
84148
answered Dec 6 at 15:42
iamklaus
84148
answered Dec 6 at 15:42
iamklaus
84148
84148
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
1
1
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
1
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
563420
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
1
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
563420
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
1
1
1
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
563420
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
563420
answered Dec 6 at 15:55
Sander van den Oord
563420
answered Dec 6 at 15:55
Sander van den Oord
563420
answered Dec 6 at 15:55
Sander van den Oord
563420
563420
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
Be careful, this sets all values to
0 after the specified index, not just the ones equal to 1. Though that's the same with some other answers too.– jpp
Dec 6 at 16:42
Be careful, this sets all values to
0 after the specified index, not just the ones equal to 1. Though that's the same with some other answers too.– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
0
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
0
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
0
0
0
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
3
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
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