python/pandas: convert month int to month name

Question

Most of the info I found was not in python>pandas>dataframe hence the question.

I want to transform an integer between 1 and 12 into an abbrieviated month name.

I have a df which looks like:

   client Month
1  sss    02
2  yyy    12
3  www    06

I want the df to look like this:

   client Month
1  sss    Feb
2  yyy    Dec
3  www    Jun

Solution 1

You can do this efficiently with combining calendar.month_abbr and df[col].apply()

import calendar
df['Month'] = df['Month'].apply(lambda x: calendar.month_abbr[x])

Solution 2

Since the abbreviated month names is the first three letters of their full names, we could first convert the Month column to datetime and then use dt.month_name() to get the full month name and finally use str.slice() method to get the first three letters, all using pandas and only in one line of code:

df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.month_name().str.slice(stop=3)

df

  Month client
0   Feb sss
1   Dec yyy
2   Jun www

Solution 3

The calendar module is useful, but calendar.month_abbr is array-like: it cannot be used directly in a vectorised fashion. For an efficient mapping, you can construct a dictionary and then use pd.Series.map:

import calendar
d = dict(enumerate(calendar.month_abbr))
df['Month'] = df['Month'].map(d)

Performance benchmarking shows a ~130x performance differential:

import calendar

d = dict(enumerate(calendar.month_abbr))
mapper = calendar.month_abbr.__getitem__

np.random.seed(0)
n = 10**5
df = pd.DataFrame({'A': np.random.randint(1, 13, n)})

%timeit df['A'].map(d)       # 7.29 ms per loop
%timeit df['A'].map(mapper)  # 946 ms per loop

Solution 4

You can do this easily with a column apply.

import pandas as pd

df = pd.DataFrame({'client':['sss', 'yyy', 'www'], 'Month': ['02', '12', '06']})

look_up = {'01': 'Jan', '02': 'Feb', '03': 'Mar', '04': 'Apr', '05': 'May',
            '06': 'Jun', '07': 'Jul', '08': 'Aug', '09': 'Sep', '10': 'Oct', '11': 'Nov', '12': 'Dec'}

df['Month'] = df['Month'].apply(lambda x: look_up[x])
df

  Month client
0   Feb    sss
1   Dec    yyy
2   Jun    www

Solution 5

def mapper(month):
   return month.strftime('%b') 

df['Month'] = df['Month'].apply(mapper)

Reference:

• http://strftime.org/

View more solutions

Comments

[jpp]

Note this solution is akin to list.__getitem__ in a Python-level loop, i.e. it does not take advantage of vectorised functionality available to Pandas. Extracting to a dictionary and then mapping is much more efficient, as per this answer.

[Nurul Akter Towhid]

I don't think it's a good idea while you have pandas own month_name() function.

[tdy]

note that .str.slice(stop=3) can be reduced to just .str[:3]