python pandas：将带参数的函数应用于系列。更新
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python pandas: apply a function with arguments to a series. Update
提问by user2988577
I would like to apply a function with argument to a pandas series: I have found two different solution of SO:
我想将带参数的函数应用于Pandas系列：我找到了两种不同的 SO 解决方案：
python pandas: apply a function with arguments to a series
and
和
Passing multiple arguments to apply (Python)
both of them rely on the use of functool.partialand they works absolutely fine. By the way the new version of Pandas support multiple argument: in any case I do not understand how does it works. Example:
它们都依赖于使用functool.partial并且它们工作得非常好。顺便说一下，新版本的 Pandas 支持多个参数：无论如何我不明白它是如何工作的。例子：
a=pd.DataFrame({'x':[1,2],'y':[10,20]})
a['x'].apply(lambda x,y: x+y, args=(100))
It exits with a:
它以以下方式退出：
TypeError: <lambda>() argument after * must be a sequence, not int
回答by TomAugspurger
The TypeErroris saying that you passed the wrong type to the lambdafunction x + y. It's expecting the argsto be a sequence, but it got an int. You may have thought that (100)was a tuple (a sequence), but in python it's the comma that makes a tuple:
该TypeError是说，你传递了错误的类型的lambda功能x + y。它期望args是一个序列，但它有一个int. 你可能认为这(100)是一个元组（一个序列），但在 python 中，逗号构成了一个元组：
In [10]: type((100))
Out[10]: int
In [11]: type((100,))
Out[11]: tuple
So change your last line to
所以将你的最后一行更改为
In [12]: a['x'].apply(lambda x, y: x + y, args=(100,))
Out[12]:
0 101
1 102
Name: x, dtype: int64
