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来完成(groupby、aggregate、apply 等)，但我发现过滤行的唯一方法是通过普通括号索引

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，因为它需要我将 df 分配给一个变量，然后才能对其值进行过滤.有没有更像下面的内容?

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

我不完全确定你想要什么，你的最后一行代码也没有帮助，但无论如何:

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

链接"过滤是通过链接"布尔索引中的条件来完成的.

"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，可以添加自己的掩码方法并使用该方法.

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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