Get month name from number
Question:
How can I get the month name from the month number?
For instance, if I have 3
, I want to return march
date.tm_month()
How to get the string march
?
Answers:
import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B")
Returns: December
Some more info on the Python doc website
[EDIT : great comment from @GiriB] You can also use %b
which returns the short notation for month name.
mydate.strftime("%b")
For the example above, it would return Dec
.
From that you can see that calendar.month_name[3]
would return March
, and the array index of 0
is the empty string, so there’s no need to worry about zero-indexing either.
import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B") # 'December'
mydate.strftime("%b") # 'dec'
import datetime
monthinteger = 4
month = datetime.date(1900, monthinteger, 1).strftime('%B')
print month
April
I’ll offer this in case (like me) you have a column of month numbers in a dataframe:
df['monthName'] = df['monthNumer'].apply(lambda x: calendar.month_name[x])
This is not so helpful if you need to just know the month name for a given number (1 – 12), as the current day doesn’t matter.
calendar.month_name[i]
or
calendar.month_abbr[i]
are more useful here.
Here is an example:
import calendar
for month_idx in range(1, 13):
print (calendar.month_name[month_idx])
print (calendar.month_abbr[month_idx])
print ("")
Sample output:
January
Jan
February
Feb
March
Mar
...
This Is What I Would Do:
from datetime import *
months = ["Unknown",
"January",
"Febuary",
"March",
"April",
"May",
"June",
"July",
"August",
"September",
"October",
"November",
"December"]
now = (datetime.now())
year = (now.year)
month = (months[now.month])
print(month)
It Outputs:
>>> September
(This Was The Real Date When I Wrote This)
I created my own function converting numbers to their corresponding month.
def month_name (number):
if number == 1:
return "January"
elif number == 2:
return "February"
elif number == 3:
return "March"
elif number == 4:
return "April"
elif number == 5:
return "May"
elif number == 6:
return "June"
elif number == 7:
return "July"
elif number == 8:
return "August"
elif number == 9:
return "September"
elif number == 10:
return "October"
elif number == 11:
return "November"
elif number == 12:
return "December"
Then I can call the function. For example:
print (month_name (12))
Outputs:
>>> December
For arbitaray range of month numbers
month_integer=range(0,100)
map(lambda x: calendar.month_name[x%12+start],month_integer)
will yield correct list. Adjust start
-parameter from where January begins in the month-integer list.
8.1. datetime — Basic date and time types — Python 2.7.17 documentation
https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
A list of all the strftime arguments. Names of months and nice stuff like formatting left zero fill. Read the full page for stuff like rules for “naive” arguments. Here is the list in brief:
%a Sun, Mon, …, Sat
%A Sunday, Monday, …, Saturday
%w Weekday as number, where 0 is Sunday
%d Day of the month 01, 02, …, 31
%b Jan, Feb, …, Dec
%B January, February, …, December
%m Month number as a zero-padded 01, 02, …, 12
%y 2 digit year zero-padded 00, 01, …, 99
%Y 4 digit Year 1970, 1988, 2001, 2013
%H Hour (24-hour clock) zero-padded 00, 01, …, 23
%I Hour (12-hour clock) zero-padded 01, 02, …, 12
%p AM or PM.
%M Minute zero-padded 00, 01, …, 59
%S Second zero-padded 00, 01, …, 59
%f Microsecond zero-padded 000000, 000001, …, 999999
%z UTC offset in the form +HHMM or -HHMM +0000, -0400, +1030
%Z Time zone name UTC, EST, CST
%j Day of the year zero-padded 001, 002, …, 366
%U Week number of the year zero padded, Days before the first Sunday are week 0
%W Week number of the year (Monday as first day)
%c Locale’s date and time representation. Tue Aug 16 21:30:00 1988
%x Locale’s date representation. 08/16/1988 (en_US)
%X Locale’s time representation. 21:30:00
%% literal ‘%’ character.
Some good answers already make use of calendar but the effect of setting the locale hasn’t been mentioned yet.
Calendar set month names according to the current locale, for exemple in French:
import locale
import calendar
locale.setlocale(locale.LC_ALL, 'fr_FR')
assert calendar.month_name[1] == 'janvier'
assert calendar.month_abbr[1] == 'jan'
If you plan on using setlocale
in your code, make sure to read the tips and caveats and extension writer sections from the documentation. The example shown here is not representative of how it should be used. In particular, from these two sections:
It is generally a bad idea to call setlocale() in some library routine, since as a side effect it affects the entire program […]
Extension modules should never call setlocale() […]
To print all months at once:
import datetime
monthint = list(range(1,13))
for X in monthint:
month = datetime.date(1900, X , 1).strftime('%B')
print(month)
This script is to show how to get calendar month abbreviations for a month variable/column in a data frame. Note that the assumption is that the values of the month column/variable are all numbers and maybe there might be some missing values.
# Import the calendar module
import calendar
# Extract month as a number from the date column
df['Month']=pd.DatetimeIndex(df['Date']).month
# Using list comprehension extract month abbreviations for each month number
df['Month_abbr']=[calendar.month_abbr[int(i)] if pd.notna(i) else i for i in df['Month']]
instead of import or downloading a new lib, you can use my manual function:
copy the code to a new .py file and run it
quick explanation:
change your_number and print month_from_number, you will get the month
your_number = 10
def num_to_month(*args, **kwargs):
num = kwargs.get("num", None)
if int(num) <= 12 and int(num) > 0:
list_of_months = {'1': 'January', '2': 'February', '3': 'March',
'4': 'April', '5': 'May', '6': 'June', '7': 'July',
'8': 'August', '9': 'September', '10': 'October',
'11': 'November', '12': 'December'}
return list_of_months[num]
else:
print('num_to_month function error: "num=' + str(num) + '"')
month_from_num = num_to_month(num=str(your_number))
print(month_from_number)
the result will be October
’01’ to ‘Jan’
from datetime import datetime
datetime.strptime('01', "%m").strftime("%b")
If you just have the month number and not a datetime instance you could built a small function to map number with label as this:
def monthname(m):
d={1:'En',2:'Fe',3:'Mr',4:'Ab',5:'My',6:'Jn',7:'Jl',8:'Ag',9:'Se',10:'Oc',11:'No',12:'Di'}
return d[m]
How can I get the month name from the month number?
For instance, if I have 3
, I want to return march
date.tm_month()
How to get the string march
?
import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B")
Returns: December
Some more info on the Python doc website
[EDIT : great comment from @GiriB] You can also use %b
which returns the short notation for month name.
mydate.strftime("%b")
For the example above, it would return Dec
.
From that you can see that calendar.month_name[3]
would return March
, and the array index of 0
is the empty string, so there’s no need to worry about zero-indexing either.
import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B") # 'December'
mydate.strftime("%b") # 'dec'
import datetime
monthinteger = 4
month = datetime.date(1900, monthinteger, 1).strftime('%B')
print month
April
I’ll offer this in case (like me) you have a column of month numbers in a dataframe:
df['monthName'] = df['monthNumer'].apply(lambda x: calendar.month_name[x])
This is not so helpful if you need to just know the month name for a given number (1 – 12), as the current day doesn’t matter.
calendar.month_name[i]
or
calendar.month_abbr[i]
are more useful here.
Here is an example:
import calendar
for month_idx in range(1, 13):
print (calendar.month_name[month_idx])
print (calendar.month_abbr[month_idx])
print ("")
Sample output:
January
Jan
February
Feb
March
Mar
...
This Is What I Would Do:
from datetime import *
months = ["Unknown",
"January",
"Febuary",
"March",
"April",
"May",
"June",
"July",
"August",
"September",
"October",
"November",
"December"]
now = (datetime.now())
year = (now.year)
month = (months[now.month])
print(month)
It Outputs:
>>> September
(This Was The Real Date When I Wrote This)
I created my own function converting numbers to their corresponding month.
def month_name (number):
if number == 1:
return "January"
elif number == 2:
return "February"
elif number == 3:
return "March"
elif number == 4:
return "April"
elif number == 5:
return "May"
elif number == 6:
return "June"
elif number == 7:
return "July"
elif number == 8:
return "August"
elif number == 9:
return "September"
elif number == 10:
return "October"
elif number == 11:
return "November"
elif number == 12:
return "December"
Then I can call the function. For example:
print (month_name (12))
Outputs:
>>> December
For arbitaray range of month numbers
month_integer=range(0,100)
map(lambda x: calendar.month_name[x%12+start],month_integer)
will yield correct list. Adjust start
-parameter from where January begins in the month-integer list.
8.1. datetime — Basic date and time types — Python 2.7.17 documentation
https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
A list of all the strftime arguments. Names of months and nice stuff like formatting left zero fill. Read the full page for stuff like rules for “naive” arguments. Here is the list in brief:
%a Sun, Mon, …, Sat
%A Sunday, Monday, …, Saturday
%w Weekday as number, where 0 is Sunday
%d Day of the month 01, 02, …, 31
%b Jan, Feb, …, Dec
%B January, February, …, December
%m Month number as a zero-padded 01, 02, …, 12
%y 2 digit year zero-padded 00, 01, …, 99
%Y 4 digit Year 1970, 1988, 2001, 2013
%H Hour (24-hour clock) zero-padded 00, 01, …, 23
%I Hour (12-hour clock) zero-padded 01, 02, …, 12
%p AM or PM.
%M Minute zero-padded 00, 01, …, 59
%S Second zero-padded 00, 01, …, 59
%f Microsecond zero-padded 000000, 000001, …, 999999
%z UTC offset in the form +HHMM or -HHMM +0000, -0400, +1030
%Z Time zone name UTC, EST, CST
%j Day of the year zero-padded 001, 002, …, 366
%U Week number of the year zero padded, Days before the first Sunday are week 0
%W Week number of the year (Monday as first day)
%c Locale’s date and time representation. Tue Aug 16 21:30:00 1988
%x Locale’s date representation. 08/16/1988 (en_US)
%X Locale’s time representation. 21:30:00
%% literal ‘%’ character.
Some good answers already make use of calendar but the effect of setting the locale hasn’t been mentioned yet.
Calendar set month names according to the current locale, for exemple in French:
import locale
import calendar
locale.setlocale(locale.LC_ALL, 'fr_FR')
assert calendar.month_name[1] == 'janvier'
assert calendar.month_abbr[1] == 'jan'
If you plan on using setlocale
in your code, make sure to read the tips and caveats and extension writer sections from the documentation. The example shown here is not representative of how it should be used. In particular, from these two sections:
It is generally a bad idea to call setlocale() in some library routine, since as a side effect it affects the entire program […]
Extension modules should never call setlocale() […]
To print all months at once:
import datetime
monthint = list(range(1,13))
for X in monthint:
month = datetime.date(1900, X , 1).strftime('%B')
print(month)
This script is to show how to get calendar month abbreviations for a month variable/column in a data frame. Note that the assumption is that the values of the month column/variable are all numbers and maybe there might be some missing values.
# Import the calendar module
import calendar
# Extract month as a number from the date column
df['Month']=pd.DatetimeIndex(df['Date']).month
# Using list comprehension extract month abbreviations for each month number
df['Month_abbr']=[calendar.month_abbr[int(i)] if pd.notna(i) else i for i in df['Month']]
instead of import or downloading a new lib, you can use my manual function:
copy the code to a new .py file and run it
quick explanation:
change your_number and print month_from_number, you will get the month
your_number = 10
def num_to_month(*args, **kwargs):
num = kwargs.get("num", None)
if int(num) <= 12 and int(num) > 0:
list_of_months = {'1': 'January', '2': 'February', '3': 'March',
'4': 'April', '5': 'May', '6': 'June', '7': 'July',
'8': 'August', '9': 'September', '10': 'October',
'11': 'November', '12': 'December'}
return list_of_months[num]
else:
print('num_to_month function error: "num=' + str(num) + '"')
month_from_num = num_to_month(num=str(your_number))
print(month_from_number)
the result will be October
’01’ to ‘Jan’
from datetime import datetime
datetime.strptime('01', "%m").strftime("%b")
If you just have the month number and not a datetime instance you could built a small function to map number with label as this:
def monthname(m):
d={1:'En',2:'Fe',3:'Mr',4:'Ab',5:'My',6:'Jn',7:'Jl',8:'Ag',9:'Se',10:'Oc',11:'No',12:'Di'}
return d[m]