大 pandas :使用运算符链接过滤DataFrame的行 [英] pandas: filter rows of DataFrame with operator chaining

问题描述

pandas中的大多数操作都可以通过操作符链接(groupby，aggregate，apply等)完成，但是我发现过滤行的唯一方法是通过常规括号索引

Most operations in pandas can be accomplished with operator chaining (groupby, aggregate, apply, etc), but the only way I've found to filter rows is via normal bracket indexing

df_filtered = df[df['column'] == value]

这并不吸引人，因为它要求我先将df分配给变量，然后才能根据其值进行过滤.还有其他类似的东西吗?

This is unappealing as it requires I assign df to a variable before being able to filter on its values. Is there something more like the following?

df_filtered = df.mask(lambda x: x['column'] == value)

推荐答案

我不确定您想要什么，最后一行代码也无济于事，但是无论如何:

I'm not entirely sure what you want, and your last line of code does not help either, but anyway:

链接"过滤是通过将布尔索引中的条件链接"完成的.

"Chained" filtering is done by "chaining" the criteria in the boolean index.

In [96]: df
Out[96]:
   A  B  C  D
a  1  4  9  1
b  4  5  0  2
c  5  5  1  0
d  1  3  9  6

In [99]: df[(df.A == 1) & (df.D == 6)]
Out[99]:
   A  B  C  D
d  1  3  9  6

如果要链接方法，可以添加自己的mask方法并使用该方法.

If you want to chain methods, you can add your own mask method and use that one.

In [90]: def mask(df, key, value):
   ....:     return df[df[key] == value]
   ....:

In [92]: pandas.DataFrame.mask = mask

In [93]: df = pandas.DataFrame(np.random.randint(0, 10, (4,4)), index=list('abcd'), columns=list('ABCD'))

In [95]: df.ix['d','A'] = df.ix['a', 'A']

In [96]: df
Out[96]:
   A  B  C  D
a  1  4  9  1
b  4  5  0  2
c  5  5  1  0
d  1  3  9  6

In [97]: df.mask('A', 1)
Out[97]:
   A  B  C  D
a  1  4  9  1
d  1  3  9  6

In [98]: df.mask('A', 1).mask('D', 6)
Out[98]:
   A  B  C  D
d  1  3  9  6

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