如果(像我一样)你在数据框架中有一列月份数字，我会提供这个:

df['monthName'] = df['monthNumer'].apply(lambda x: calendar.month_name[x])

我创建了自己的函数，将数字转换为对应的月份。

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"

然后我可以调用函数。例如:

print (month_name (12))

输出:

>>> December

import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B") # 'December'
mydate.strftime("%b") # 'dec'

一些好的答案已经使用了日历，但是设置语言环境的效果还没有被提及。

Calendar根据当前地区设置月份名称，例如在法语中:

import locale
import calendar

locale.setlocale(locale.LC_ALL, 'fr_FR')

assert calendar.month_name[1] == 'janvier'
assert calendar.month_abbr[1] == 'jan'

如果您计划在代码中使用setlocale，请确保阅读文档中的提示和警告以及扩展编写器部分。这里显示的示例不能代表它应该如何使用。特别是这两个部分:

在某些库例程中调用setlocale()通常不是一个好主意，因为它会影响整个程序[…] 扩展模块永远不应该调用setlocale()[…]

这就是我要做的:

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)

输出:

>>> September

(这是我写这篇文章的真实日期)

import datetime

monthinteger = 4

month = datetime.date(1900, monthinteger, 1).strftime('%B')

print month

4月