Mixing
Python pandas dataframe change content based on string length
2
I have a dataframe like below.
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
so if you look at the row 3,4,5 the length of the ids are less than 6
id
111111
123456
12345
234
12
987654
I want to convert it to below(Basically append zeroes in the front to make it a length of six)
id
111111
123456
012345
000234
000012
987654
python pandas dataframe
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
add a comment |
2
I have a dataframe like below.
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
so if you look at the row 3,4,5 the length of the ids are less than 6
id
111111
123456
12345
234
12
987654
I want to convert it to below(Basically append zeroes in the front to make it a length of six)
id
111111
123456
012345
000234
000012
987654
python pandas dataframe
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41
add a comment |
2
2
2
I have a dataframe like below.
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
so if you look at the row 3,4,5 the length of the ids are less than 6
id
111111
123456
12345
234
12
987654
I want to convert it to below(Basically append zeroes in the front to make it a length of six)
id
111111
123456
012345
000234
000012
987654
python pandas dataframe
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
I have a dataframe like below.
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
so if you look at the row 3,4,5 the length of the ids are less than 6
id
111111
123456
12345
234
12
987654
I want to convert it to below(Basically append zeroes in the front to make it a length of six)
id
111111
123456
012345
000234
000012
987654
python pandas dataframe
python pandas dataframe
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
asked Jan 2 at 2:38
Sandhya ThotakuraSandhya Thotakura
3369
3369
Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41
add a comment |
Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41
Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41
Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41
add a comment |
3 Answers
3
active
oldest
votes
2
Using
df.id.astype(str).str.pad(6,'left','0')
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
add a comment |
3
Try zfill, first convert integers to string dtype then use string accessor then zfill:
df['id'].astype(str).str.zfill(6)
Output:
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
add a comment |
1
You have to change the datatype to string or else the initial zeroes will be trimmed.
Below is my solution. Hope it helps
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
df = df.astype(str)
df['id'] = df['id'].apply(lambda x: x if len(x)==6 else '0'*(6-len(x))+x)
df:
id
111111
123456
012345
000234
000012
987654
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Using
df.id.astype(str).str.pad(6,'left','0')
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
add a comment |
2
Using
df.id.astype(str).str.pad(6,'left','0')
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
add a comment |
2
2
2
Using
df.id.astype(str).str.pad(6,'left','0')
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
Using
df.id.astype(str).str.pad(6,'left','0')
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
answered Jan 2 at 2:45
Wen-BenWen-Ben
118k83469
118k83469
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
add a comment |
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
1
1
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
@ScottBoston done with that : -)
– Wen-Ben
Jan 2 at 2:47
Very kind of you.
– Scott Boston
Jan 2 at 2:48
Very kind of you.
– Scott Boston
Jan 2 at 2:48
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
@SandhyaThotakura yw :_ happy new year
– Wen-Ben
Jan 2 at 2:55
add a comment |
3
Try zfill, first convert integers to string dtype then use string accessor then zfill:
df['id'].astype(str).str.zfill(6)
Output:
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
add a comment |
3
Try zfill, first convert integers to string dtype then use string accessor then zfill:
df['id'].astype(str).str.zfill(6)
Output:
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
add a comment |
3
3
3
Try zfill, first convert integers to string dtype then use string accessor then zfill:
df['id'].astype(str).str.zfill(6)
Output:
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
Try zfill, first convert integers to string dtype then use string accessor then zfill:
df['id'].astype(str).str.zfill(6)
Output:
0 111111
1 123456
2 012345
3 000234
4 000012
5 987654
Name: id, dtype: object
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
answered Jan 2 at 2:45
Scott BostonScott Boston
56.7k73158
56.7k73158
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
add a comment |
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
1
1
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
Thanks for the answer. But I will go with W-B as I feel his answer is more robust. Thanks for teaching me the concept of zfill(int)
– Sandhya Thotakura
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
@SandhyaThotakura Happy coding!
– Scott Boston
Jan 2 at 2:53
add a comment |
1
You have to change the datatype to string or else the initial zeroes will be trimmed.
Below is my solution. Hope it helps
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
df = df.astype(str)
df['id'] = df['id'].apply(lambda x: x if len(x)==6 else '0'*(6-len(x))+x)
df:
id
111111
123456
012345
000234
000012
987654
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
add a comment |
1
You have to change the datatype to string or else the initial zeroes will be trimmed.
Below is my solution. Hope it helps
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
df = df.astype(str)
df['id'] = df['id'].apply(lambda x: x if len(x)==6 else '0'*(6-len(x))+x)
df:
id
111111
123456
012345
000234
000012
987654
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
add a comment |
1
1
1
You have to change the datatype to string or else the initial zeroes will be trimmed.
Below is my solution. Hope it helps
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
df = df.astype(str)
df['id'] = df['id'].apply(lambda x: x if len(x)==6 else '0'*(6-len(x))+x)
df:
id
111111
123456
012345
000234
000012
987654
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
You have to change the datatype to string or else the initial zeroes will be trimmed.
Below is my solution. Hope it helps
df = pd.DataFrame([111111,123456,12345,234,12,987654],columns=['id'])
df = df.astype(str)
df['id'] = df['id'].apply(lambda x: x if len(x)==6 else '0'*(6-len(x))+x)
df:
id
111111
123456
012345
000234
000012
987654
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
answered Jan 2 at 2:40
Bhanu TezBhanu Tez
41519
41519
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
add a comment |
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote.
– Sandhya Thotakura
Jan 2 at 2:55
add a comment |
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Perhaps take a look at stackoverflow.com/questions/134934/…
– sfjac
Jan 2 at 2:41