如何从数字列中提取数字的任何子集

Question

I have an integer column(int64) in a dataframe with values as 20170811. (yyyymmdd) Now i need to extract 08 only and store it as a new column.

df['key'].floordiv(10000) 

gives me 2017. But i wanted 08 - how to get it ?

Answer 1

New Answer (as requested in comments)

Converting to Datetime has several advantages if you want to format your datetime string. In order to do so you can use DataFrame.dt.strftime . For more information on how to format strings and create custmo formats take a look at this .

import pandas as pd

df = pd.DataFrame({'key': ['20181201', '20180302', '20180403']})

pd.to_datetime(df.key).dt.strftime('%b')

This will output:

0    Dec
1    Mar
2    Apr
Name: key, dtype: object

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Old Answer

What you can do is take the module of the value, divide it by 100 and drop the decimals:

import pandas as pd

df = pd.DataFrame({'key': ['20181201', '20180302', '20180403']}).astype(int)

df['key'].map(lambda x: int((x % 10000) / 100))

Which outputs:

0    12
1     3
2     4
Name: key, dtype: int64

In case you have strings, you could convert it to Datetime objects and simply access month :

import pandas as pd

df = pd.DataFrame({'key': ['20181201', '20180302', '20180403']})

pd.to_datetime(df.key).map(lambda x: x.month)

giving you the same output.

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Answer 2

Or use:

df['key']=df.astype(str)['key'].map(lambda x: x[4:6]).astype(int)

As you said to @Stefan, do:

import calendar
df['key']=df.astype(str)['key'].map(lambda x: x[4:6]).astype(int).apply(lambda x: calendar.month_name[int(x)-1])

Or apply :

df['key']=df.astype(str)['key'].apply(lambda x: x[4:6]).astype(int)

As you said to @Stefan, do:

import calendar
df['key']=df.astype(str)['key'].apply(lambda x: x[4:6]).astype(int).apply(lambda x: calendar.month_name[int(x)-1])

Answer 3

Probably the most robust way is:

import datetime
import pandas as pd

df = pd.DataFrame({'date': [20180201, 20180302, 20180403]})
df['month'] = pd.to_datetime(df['date'].astype(str), format='%Y%m%d').map(lambda x: x.strftime('%m'))

If you wanted the df['month'] to be integer, just cast it with col.astype(int) .

Edit: If you wanted your month in a format of Apr, May, etc., use x.strftime('%b') . You may want to look at strftime documentation .

Answer 4

You can convert your series to a string and then slice using Pandas str methods:

df = pd.DataFrame({'date': [20180201, 20180302, 20180403]})

df['key'] = df['date'].astype(str).str[4:6]

print(df)

       date key
0  20180201  02
1  20180302  03
2  20180403  04

A much better alternative is to convert to datetime and extract months as integers:

df['key'] = pd.to_datetime(df['date'].astype(str)).dt.month

print(df)

       date  key
0  20180201    2
1  20180302    3
2  20180403    4