设置数据框之间的列差异 [英] Set differences on columns between dataframes

问题描述

注意:这个问题的灵感来自另一篇文章中讨论的想法:Pandas 中的 DataFrame 代数

Note: This question is inspired by the ideas discussed in this other post: DataFrame algebra in Pandas

假设我有两个数据框 A 和 B 并且对于某些列 col_name，它们的值为:

Say I have two dataframes A and B and that for some column col_name, their values are:

A[col_name]   |  B[col_name]  
--------------| ------------
1             |  3
2             |  4
3             |  5
4             |  6

我想根据 col_name 计算 A 和 B 之间的集合差.这个操作的结果应该是:

I want to compute the set difference between A and B based on col_name. The result of this operation should be:

A 的行，其中 A[col_name] 与 B[col_name] 中的任何条目都不匹配.

The rows of A where A[col_name] didn't match any entries in B[col_name].

以下是上述示例的结果(也显示了 A 的其他列):

Below is the result for the above example (showing other columns of A as well):

A[col_name] | A[other_column_1] | A[other_column_2]  
------------+-------------------|------------------ 
1           |    'foo'          |  'xyz'            ....
2           |    'bar'          |  'abc'

请记住，A[col_name] 和 B[col_name] 中的某些条目可能包含值 np.NaN.我想将这些条目视为未定义但不同的，即集合差异应该返回它们.

Keep in mind that some entries in A[col_name] and B[col_name] could hold the value np.NaN. I would like to treat those entries as undefined BUT different, i.e. the set difference should return them.

我怎样才能在 Pandas 中做到这一点?(概括为多列上的差异也很好)

How can I do this in Pandas? (generalizing to a difference on multiple columns would be great as well)

推荐答案

一种方法是使用 Series isin 方法:

One way is to use the Series isin method:

In [11]: df1 = pd.DataFrame([[1, 'foo'], [2, 'bar'], [3, 'meh'], [4, 'baz']], columns = ['A', 'B'])

In [12]: df2 = pd.DataFrame([[3, 'a'], [4, 'b']], columns = ['A', 'C'])

现在可以检查df1['A']中的每一项是否在df2['A']中:

Now you can check whether each item in df1['A'] is in of df2['A']:

In [13]: df1['A'].isin(df2['A'])
Out[13]:
0    False
1    False
2     True
3     True
Name: A, dtype: bool

In [14]: df1[~df1['A'].isin(df2['A'])]  # not in df2['A']
Out[14]:
   A    B
0  1  foo
1  2  bar

我认为这也符合您对 NaN 的要求:

I think this does what you want for NaNs too:

In [21]: df1 = pd.DataFrame([[1, 'foo'], [np.nan, 'bar'], [3, 'meh'], [np.nan, 'baz']], columns = ['A', 'B'])

In [22]: df2 = pd.DataFrame([[3], [np.nan]], columns = ['A'])

In [23]: df1[~df1['A'].isin(df2['A'])]
Out[23]:
    A     B
0 1.0   foo
1 NaN   bar
3 NaN   baz

注意:对于大型框架，可能值得将这些列设为索引(以按照 另一个问题).

合并两个或多个列的一种方法是使用虚拟列:

One way to merge on two or more columns is to use a dummy column:

In [31]: df1 = pd.DataFrame([[1, 'foo'], [np.nan, 'bar'], [4, 'meh'], [np.nan, 'eurgh']], columns = ['A', 'B'])

In [32]: df2 = pd.DataFrame([[np.nan, 'bar'], [4, 'meh']], columns = ['A', 'B'])

In [33]: cols = ['A', 'B']

In [34]: df2['dummy'] = df2[cols].isnull().any(1)  # rows with NaNs in cols will be True

In [35]: merged = df1.merge(df2[cols + ['dummy']], how='left')

In [36]: merged
Out[36]:
    A      B  dummy
0   1    foo    NaN
1 NaN    bar   True
2   4    meh  False
3 NaN  eurgh    NaN

布尔值存在于 df2 中，True 在合并列之一中具有 NaN.按照您的规范，我们应该删除那些错误的:

The booleans were present in df2, the True has an NaN in one of the merging columns. Following your spec, we should drop those which are False:

In [37]: merged.loc[merged.dummy != False, df1.columns]
Out[37]:
    A      B
0   1    foo
1 NaN    bar
3 NaN  eurgh

不优雅.

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