将前导零添加到 Pandas Dataframe 中的字符串

Question

I have a pandas data frame where the first 3 columns are strings:

         ID        text1    text 2
0       2345656     blah      blah
1          3456     blah      blah
2        541304     blah      blah        
3        201306       hi      blah        
4   12313201308    hello      blah         

I want to add leading zeros to the ID:

                ID    text1    text 2
0  000000002345656     blah      blah
1  000000000003456     blah      blah
2  000000000541304     blah      blah        
3  000000000201306       hi      blah        
4  000012313201308    hello      blah 

I have tried:

df['ID'] = df.ID.zfill(15)
df['ID'] = '{0:0>15}'.format(df['ID'])

Answer 1

str attribute contains most of the methods in string.

df['ID'] = df['ID'].str.zfill(15)

See more: http://pandas.pydata.org/pandas-docs/stable/text.html

Answer 2

Try:

df['ID'] = df['ID'].apply(lambda x: '{0:0>15}'.format(x))

or even

df['ID'] = df['ID'].apply(lambda x: x.zfill(15))

Answer 3

It can be achieved with a single line while initialization. Just use converters argument.

df = pd.read_excel('filename.xlsx', converters={'ID': '{:0>15}'.format})

so you'll reduce the code length by half :)

PS: read_csv have this argument as well.

Answer 4

With Python 3.6+, you can also use f-strings:

df['ID'] = df['ID'].map(lambda x: f'{x:0>15}')

Performance is comparable or slightly worse versus df['ID'].map('{:0>15}'.format) . On the other hand, f-strings permit more complex output, and you can use them more efficiently via a list comprehension.

Performance benchmarking

# Python 3.6.0, Pandas 0.19.2

df = pd.concat([df]*1000)

%timeit df['ID'].map('{:0>15}'.format)                  # 4.06 ms per loop
%timeit df['ID'].map(lambda x: f'{x:0>15}')             # 5.46 ms per loop
%timeit df['ID'].astype(str).str.zfill(15)              # 18.6 ms per loop

%timeit list(map('{:0>15}'.format, df['ID'].values))    # 7.91 ms per loop
%timeit ['{:0>15}'.format(x) for x in df['ID'].values]  # 7.63 ms per loop
%timeit [f'{x:0>15}' for x in df['ID'].values]          # 4.87 ms per loop
%timeit [str(x).zfill(15) for x in df['ID'].values]     # 21.2 ms per loop

# check results are the same
x = df['ID'].map('{:0>15}'.format)
y = df['ID'].map(lambda x: f'{x:0>15}')
z = df['ID'].astype(str).str.zfill(15)

assert (x == y).all() and (x == z).all()

Answer 5

If you are encountering the error:

Pandas error: Can only use .str accessor with string values, which use np.object_ dtype in pandas

df['ID'] = df['ID'].astype(str).str.zfill(15)

Answer 6

If you want a more customizable solution to this problem, you can try pandas.Series.str.pad

df['ID'] = df['ID'].astype(str).str.pad(15, side='left', fillchar='0')

str.zfill(n) is a special case equivalent to str.pad(n, side='left', fillchar='0')

Answer 7

刚刚为我工作：

df['ID']= df['ID'].str.rjust(15,'0')