Python：在 Pandas lambda 表达式中使用函数
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Python: use a function in pandas lambda expression
提问by Edamame
I have the following code, trying to find the hour of the 'Dates' column in a data frame:
我有以下代码，试图在数据框中查找“日期”列的小时数：
print(df['Dates'].head(3))
df['hour'] = df.apply(lambda x: find_hour(x['Dates']), axis=1)
def find_hour(self, input):
return input[11:13].astype(float)
where the print(df['Dates'].head(3))looks like:
其中print(df['Dates'].head(3))的样子：
0 2015-05-13 23:53:00
1 2015-05-13 23:53:00
2 2015-05-13 23:33:00
However, I got the following error:
但是，我收到以下错误：
df['hour'] = df.apply(lambda x: find_hour(x['Dates']), axis=1)
NameError: ("global name 'find_hour' is not defined", u'occurred at index 0')
Does anyone know what I missed? Thanks!
有谁知道我错过了什么？谢谢！
Note that if I put the function directly in the lambda line like below, everything works fine:
请注意，如果我将函数直接放在 lambda 行中，如下所示，一切正常：
df['hour'] = df.apply(lambda x: x['Dates'][11:13], axis=1).astype(float)
回答by zondo
You are trying to use find_hourbefore it has yet been defined. You just need to switch things around:
您正在尝试find_hour在尚未定义之前使用。你只需要改变事情：
def find_hour(self, input):
return input[11:13].astype(float)
print(df['Dates'].head(3))
df['hour'] = df.apply(lambda x: find_hour(x['Dates']), axis=1)
Edit: Padraic has pointed out a very important point: find_hour()is defined as taking two arguments, selfand input, but you are giving it only one. You should define find_hour()as def find_hour(input):except that defining the argument as inputshadows the built-in function. You might consider renaming it to something a little more descriptive.
编辑：Padraic 指出了一个非常重要的点：find_hour()被定义为采用两个参数, selfand input，但你只给它一个。您应该定义find_hour()为def find_hour(input):除了将参数定义为input隐藏内置函数之外。您可能会考虑将其重命名为更具描述性的名称。
回答by MaxU
what is wrong with old good .dt.hour?
old good 有.dt.hour什么问题？
In [202]: df
Out[202]:
Date
0 2015-05-13 23:53:00
1 2015-05-13 23:53:00
2 2015-05-13 23:33:00
In [217]: df['hour'] = df.Date.dt.hour
In [218]: df
Out[218]:
Date hour
0 2015-05-13 23:53:00 23
1 2015-05-13 23:53:00 23
2 2015-05-13 23:33:00 23
and if your Datecolumn is of string type you may want to convert it to datetime first:
如果您的Date列是字符串类型，您可能希望先将其转换为日期时间：
df.Date = pd.to_datetime(df.Date)
or just:
要不就：
df['hour'] = int(df.Date.str[11:13])
