如何从月份编号中获得月份名称?
例如,如果我有3,我想返回march
date.tm_month()
如何获得弦乐进行曲?
如何从月份编号中获得月份名称?
例如,如果我有3,我想返回march
date.tm_month()
如何获得弦乐进行曲?
当前回答
日历火
从那里你可以看到那个日历。month_name[3]将返回March,数组索引为0是空字符串,因此也不需要担心零索引。
其他回答
datetime -基本日期和时间类型- Python文档
所有strftime格式代码的列表。月份的名称,以及格式化左零填充之类的好东西。阅读整页的内容,比如“天真”参数的规则。以下是清单的简要内容:
%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.
import datetime
mydate = datetime.datetime.now()
mydate.strftime("%B") # 'December'
mydate.strftime("%b") # 'dec'
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)
这个脚本展示了如何获得数据帧中月份变量/列的日历月缩写。注意,假设month列/变量的值都是数字,可能会有一些缺失的值。
# 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']]
对于任意范围的月份数字
month_integer=range(0,100)
map(lambda x: calendar.month_name[x%12+start],month_integer)
将产生正确的列表。从月份-整数列表中一月份开始的位置调整start-parameter。