Python pandas dataframe change content based on string length

Question

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

I want to convert it to below(Basically append zeroes in the front to make it a length of six)

Perhaps take a look at https://stackoverflow.com/questions/134934/display-number-with-leading-zeros — sfjac, Jan 02 '19 at 02:41

score 3 · Accepted Answer · answered Jan 02 '19 at 02:45

3

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 02 '19 at 02:45

BENY

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234

Very kind of you. – Scott Boston Jan 02 '19 at 02:48
@SandhyaThotakura yw :_ happy new year – BENY Jan 02 '19 at 02:55

score 3 · Answer 2 · answered Jan 02 '19 at 02:45

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 02 '19 at 02:45

Scott Boston

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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 02 '19 at 02:53
@SandhyaThotakura Happy coding! – Scott Boston Jan 02 '19 at 02:53

score 0 · Answer 3 · answered Jan 02 '19 at 02:40

0

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 02 '19 at 02:40

Bhanu Tez

296
2
14

Thanks Bhanu for your answer. You are quick one to answer my question. so 1 upvote. – Sandhya Thotakura Jan 02 '19 at 02:55

Python pandas dataframe change content based on string length

3 Answers3