熊猫：如何对单个列使用apply（）函数？

I have a pandas data frame with two columns. I need to change the values of the first column without affecting the second one and get back the whole data frame with just first column values changed. How can I do that using apply in pandas?

Given a sample dataframe df as:

a,b
1,2
2,3
3,4
4,5

what you want is:

df['a'] = df['a'].apply(lambda x: x + 1)

that returns:

   a  b
0  2  2
1  3  3
2  4  4
3  5  5

For a single column better to use map(), like this:

df = pd.DataFrame([{'a': 15, 'b': 15, 'c': 5}, {'a': 20, 'b': 10, 'c': 7}, {'a': 25, 'b': 30, 'c': 9}])

    a   b  c
0  15  15  5
1  20  10  7
2  25  30  9



df['a'] = df['a'].map(lambda a: a / 2.)

      a   b  c
0   7.5  15  5
1  10.0  10  7
2  12.5  30  9

You don’t need a function at all. You can work on a whole column directly.

Example data:

>>> df = pd.DataFrame({'a': [100, 1000], 'b': [200, 2000], 'c': [300, 3000]})
>>> df

      a     b     c
0   100   200   300
1  1000  2000  3000

Half all the values in column a:

>>> df.a = df.a / 2
>>> df

     a     b     c
0   50   200   300
1  500  2000  3000

Although the given responses are correct, they modify the initial data frame, which is not always desirable (and, given the OP asked for examples “using apply“, it might be they wanted a version that returns a new data frame, as apply does).

This is possible using : it is valid to assign to existing columns, as the documentation states (emphasis is mine):

Assign new columns to a DataFrame.

Returns a new object with all original columns in addition to new ones. Existing columns that are re-assigned will be overwritten.

In short:

In [1]: import pandas as pd

In [2]: df = pd.DataFrame([{'a': 15, 'b': 15, 'c': 5}, {'a': 20, 'b': 10, 'c': 7}, {'a': 25, 'b': 30, 'c': 9}])

In [3]: df.assign(a=lambda df: df.a / 2)
Out[3]: 
      a   b  c
0   7.5  15  5
1  10.0  10  7
2  12.5  30  9

In [4]: df
Out[4]: 
    a   b  c
0  15  15  5
1  20  10  7
2  25  30  9

Note that the function will be passed the whole dataframe, not only the column you want to modify, so you will need to make sure you select the right column in your lambda.

If you are really concerned about the execution speed of your apply function and you have a huge dataset to work on, you could use swifter to make faster execution, here is an example for swifter on pandas dataframe:

import pandas as pd
import swifter

def fnc(m):
    return m*3+4

df = pd.DataFrame({"m": [1,2,3,4,5,6], "c": [1,1,1,1,1,1], "x":[5,3,6,2,6,1]})

# apply a self created function to a single column in pandas
df["y"] = df.m.swifter.apply(fnc)

This will enable your all CPU cores to compute the result hence it will be much faster than normal apply functions. Try and let me know if it become useful for you.

Let me try a complex computation using datetime and considering nulls or empty spaces. I am reducing 30 years on a datetime column and using apply method as well as lambda and converting datetime format. Line if x != '' else x will take care of all empty spaces or nulls accordingly.

df['Date'] = df['Date'].fillna('')
df['Date'] = df['Date'].apply(lambda x : ((datetime.datetime.strptime(str(x), '%m/%d/%Y') - datetime.timedelta(days=30*365)).strftime('%Y%m%d')) if x != '' else x)