带有和不带有lambda的pandas apply() [英] pandas apply() with and without lambda

问题描述

当通过lambda调用pandas apply()而不是调用函数时，规则/过程是什么?下面的例子.显然没有lambda，整个序列(df [column name])将传递给"test"函数，该函数会在尝试对序列执行布尔运算时引发错误.

What is the rule/process when a function is called with pandas apply() through lambda vs. not? Examples below. Without lambda apparently, the entire series ( df[column name] ) is passed to the "test" function which throws an error trying to do a boolean operation on a series.

如果通过lambda调用了相同的函数，则它将起作用.遍历每行，每行以"x"进行迭代，并且df [列名]返回当前行中该列的单个值.

If the same function is called via lambda it works. Iteration over each row with each passed as "x" and the df[ column name ] returns a single value for that column in the current row.

这就像lambda正在移除尺寸.有人对此有解释或指向特定文档吗?谢谢.

It's like lambda is removing a dimension. Anyone have an explanation or point to the specific doc on this? Thanks.

带有lambda的示例1可以正常运行

print("probPredDF columns:", probPredDF.columns)

def test( x, y):
    if x==y:
        r = 'equal'
    else:
        r = 'not equal'
    return r    

probPredDF.apply( lambda x: test( x['yTest'], x[ 'yPred']), axis=1 ).head()

示例1输出

probPredDF columns: Index([0, 1, 'yPred', 'yTest'], dtype='object')

Out[215]:
0    equal
1    equal
2    equal
3    equal
4    equal
dtype: object

没有lambda的示例2，在系列错误时引发布尔运算

print("probPredDF columns:", probPredDF.columns)

def test( x, y):
    if x==y:
        r = 'equal'
    else:
        r = 'not equal'
    return r    

probPredDF.apply( test( probPredDF['yTest'], probPredDF[ 'yPred']), axis=1 ).head()

示例2输出

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

推荐答案

lambda没有什么神奇之处.它们是一个参数中的函数，可以内联定义，并且没有名称.您可以在需要使用lambda的函数中使用该函数，但是该函数还需要采用一个参数.您需要做类似...

There is nothing magic about a lambda. They are functions in one parameter, that can be defined inline, and do not have a name. You can use a function where a lambda is expected, but the function will need to also take one parameter. You need to do something like...

将其定义为:

def wrapper(x):
    return test(x['yTest'], x['yPred'])

用作:

probPredDF.apply(wrapper, axis=1)

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