python/pandas：将月份整数转换为月份名称
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python/pandas: convert month int to month name
提问by Boosted_d16
Most of the info I found was not in python>pandas>dataframe hence the question.
我发现的大部分信息不在 python>pandas>dataframe 中，因此是问题。
I want to transform an integer between 1 and 12 into an abbrieviated month name.
我想将 1 到 12 之间的整数转换为缩写的月份名称。
I have a df which looks like:
我有一个 df 看起来像：
client Month
1 sss 02
2 yyy 12
3 www 06
I want the df to look like this:
我希望 df 看起来像这样：
client Month
1 sss Feb
2 yyy Dec
3 www Jun
回答by EoinS
You can do this efficiently with combining calendar.month_abbrand df[col].apply()
您可以通过组合calendar.month_abbr和df[col].apply()
import calendar
df['Month'] = df['Month'].apply(lambda x: calendar.month_abbr[x])
回答by today
Since the abbreviated month names is the first three letters of their full names, we could first convert the Monthcolumn to datetimeand 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:
由于月份缩写是他们全名的前三个字母，我们可以先将Month列转换为datetime然后使用dt.month_name()得到完整的月份名称，最后使用str.slice()方法获得前三个字母，全部使用pandas并且只在一行代码：
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
回答by jpp
The calendarmodule is useful, but calendar.month_abbris 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:
该calendar模块很有用，但calendar.month_abbr类似于数组：它不能以矢量化方式直接使用。为了有效的映射，您可以构造一个字典，然后使用pd.Series.map：
import calendar
d = dict(enumerate(calendar.month_abbr))
df['Month'] = df['Month'].map(d)
Performance benchmarking shows a ~130x performance differential:
性能基准测试显示约 130 倍的性能差异：
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
回答by Suhas_Pote
def mapper(month):
return month.strftime('%b')
df['Month'] = df['Month'].apply(mapper)
Reference:
参考：
回答by andrew
You can do this easily with a column apply.
您可以使用列应用轻松完成此操作。
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
回答by pekapa
One way of doing that is with the applymethod in the dataframe but, to do that, you need a map to convert the months. You could either do that with a function / dictionary or with Python's own datetime.
一种方法是使用apply数据框中的方法，但是，要做到这一点，您需要一个地图来转换月份。您可以使用函数/字典或 Python 自己的日期时间来做到这一点。
With the datetime it would be something like:
随着日期时间，它会是这样的：
def mapper(month):
date = datetime.datetime(2000, month, 1) # You need a dateobject with the proper month
return date.strftime('%b') # %b returns the months abbreviation, other options [here][1]
df['Month'].apply(mapper)
In a simillar way, you could build your own map for custom names. It would look like this:
以类似的方式，您可以为自定义名称构建自己的地图。它看起来像这样：
months_map = {01: 'Jan', 02: 'Feb'}
def mapper(month):
return months_map[month]
Obviously, you don't need to define this functions explicitly and could use a lambdadirectly in the apply method.
显然，您不需要显式定义此函数，可以lambda直接在 apply 方法中使用 a 。
回答by Vagee
Use strptimeand lambdafunction for this:
用途strptime和lambda功能：
from time import strptime
df['Month'] = df['Month'].apply(lambda x: strptime(x,'%b').tm_mon)
回答by Heather
Having tested all of these on a large dataset, I have found the following to be fastest:
在大型数据集上测试了所有这些之后，我发现以下是最快的：
import calendar
def month_mapping():
# I'm lazy so I have a stash of functions already written so
# I don't have to write them out every time. This returns the
# {1:'Jan'....12:'Dec'} dict in the laziest way...
abbrevs = {}
for month in range (1, 13):
abbrevs[month] = calendar.month_abbr[month]
return abbrevs
abbrevs = month_mapping()
df['Month Abbrev'} = df['Date Col'].dt.month.map(mapping)
回答by Nurul Akter Towhid
You can use Pandas month_name()function. Example:
您可以使用 Pandasmonth_name()功能。例子：
>>> idx = pd.date_range(start='2018-01', freq='M', periods=3)
>>> idx
DatetimeIndex(['2018-01-31', '2018-02-28', '2018-03-31'],
dtype='datetime64[ns]', freq='M')
>>> idx.month_name()
Index(['January', 'February', 'March'], dtype='object')
For more detail visit this link.
有关更多详细信息，请访问此链接。
