CODEWITHSUNDEEP
pythonpandasstringseries
Raphadasilva
Raphadasilva

Reputation: 617

Pandas : update int Series with one or several 0

I've got a pandas Serie in a DF, like this one :

DF=pd.DataFrame([1,2,58,99,123,256],columns=["ID"])

    ID
0   1
1   2
2   58
3   99
4   123
5   256

I'm trying to modify "ID" column, in order to have something like this (always three characters, so one or two zeros for numbers with length<3) :

    ID
0   "001"
1   "002"
2   "058"
3   "099"
4   "123"
5   "256"

I tried to achieve this with a list comprenhension, astype(str), replace() and str.len(), like this :

DF["ID"]=[x.replace(x, i*"0"+x) for x,i in (DF["ID"].astype(str),int(DF[DF["ID"]].astype(str).len())-3)]

But then I had this Error

KeyError: "None of [Int64Index([1, 2, 58, 99, 123, 256], dtype='int64')] are in the [columns]"

How could I fix it ?

Upvotes: 1

Views: 80

Answers (3)

timgeb
timgeb

Reputation: 78780

In my opinion the most idiomatic way is:

>>> DF['ID'] = DF['ID'].astype(str).str.rjust(3, '0')                           
>>> DF                                                                          
    ID
0  001
1  002
2  058
3  099
4  123
5  256

Upvotes: 0

piterbarg
piterbarg

Reputation: 8219

you an use f-strings with formatting

DF['ID'] = DF['ID'].apply(lambda v: f'{v:03d}')

output (these are string btw)


    ID
0   001
1   002
2   058
3   099
4   123
5   256

Upvotes: 0

BossGman
BossGman

Reputation: 78

The error you are getting is because you are trying to pass the values in the your 'ID' series as if it were a list of column labels. There is a much simpler way of doing this using pandas apply and the str.rjust() method setting the width to 3 to pad the strings to 3 characters:

df = pd.DataFrame(dict(ID=[1,2,58,99,123,256]))

df['ID'] = df['ID'].apply(lambda x: str(x).rjust(3,'0'))

Outputs:

    ID
0   "001"
1   "002"
2   "058"
3   "099"
4   "123"
5   "256"

Upvotes: 1

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