pandas 在熊猫数据框中的每一行中查找非零值的列索引集
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find the set of column indices for non-zero values in each row in pandas' data frame
提问by Qiang Li
Is there a good way to find the set of column indices for non-zero values in each row in pandas' data frame? Do I have to traverse the data frame row-by-row?
有没有一种好方法可以在Pandas数据框中的每一行中找到非零值的列索引集？我是否必须逐行遍历数据框？
For example, the data frame is
例如，数据框是
c1 c2 c3 c4 c5 c6 c7 c8 c9
1 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 2 1 1 1 1 1 0 2
1 5 5 0 0 1 0 4 6
4 3 0 1 1 1 1 5 10
3 5 2 4 1 2 2 1 3
6 4 0 1 0 0 0 0 0
3 9 1 0 1 0 2 1 0
The output is expected to be
输出预计为
['c1','c2']
['c1']
['c2']
...
采纳答案by Younggun Kim
It seems you have to traverse the DataFrame by row.
看来您必须逐行遍历 DataFrame。
cols = df.columns
bt = df.apply(lambda x: x > 0)
bt.apply(lambda x: list(cols[x.values]), axis=1)
and you will get:
你会得到：
0 [c1, c2]
1 [c1]
2 [c2]
3 [c1]
4 [c2]
5 []
6 [c2, c3, c4, c5, c6, c7, c9]
7 [c1, c2, c3, c6, c8, c9]
8 [c1, c2, c4, c5, c6, c7, c8, c9]
9 [c1, c2, c3, c4, c5, c6, c7, c8, c9]
10 [c1, c2, c4]
11 [c1, c2, c3, c5, c7, c8]
dtype: object
If performance is matter, try to pass raw=Trueto boolean DataFrame creation like below:
如果性能很重要，请尝试传递raw=True给布尔数据帧创建，如下所示：
%timeit df.apply(lambda x: x > 0, raw=True).apply(lambda x: list(cols[x.values]), axis=1)
1000 loops, best of 3: 812 μs per loop
It brings you a better performance gain. Following is raw=False(which is default) result:
它为您带来更好的性能增益。以下是raw=False（这是默认的）结果：
%timeit df.apply(lambda x: x > 0).apply(lambda x: list(cols[x.values]), axis=1)
100 loops, best of 3: 2.59 ms per loop
回答by Dickster
How about this approach?
这种方法怎么样？
#create a True / False data frame
df_boolean = df>0
#a little helper method that uses boolean slicing internally
def bar(x,columns):
return ','.join(list(columns[x]))
#use an apply along the column axis
df_boolean['result'] = df_boolean.apply(lambda x: bar(x,df_boolean.columns),axis=1)
# filter out the empty "rows" adn grab the result column
df_result = df_boolean[df_boolean['result'] != '']['result']
#append an axis, just so each line will will output a list
lst_result = df_result.values[:,np.newaxis]
print '\n'.join([ str(myelement) for myelement in lst_result])
and this produces:
这会产生：
['c1,c2']
['c1']
['c2']
['c1']
['c2']
['c2,c3,c4,c5,c6,c7,c9']
['c1,c2,c3,c6,c8,c9']
['c1,c2,c4,c5,c6,c7,c8,c9']
['c1,c2,c3,c4,c5,c6,c7,c8,c9']
['c1,c2,c4']
['c1,c2,c3,c5,c7,c8']
回答by Andy Hayden
Potentially a better data structure (rather than a Series of lists) is to stack:
潜在更好的数据结构（而不是一系列列表）是堆栈：
In [11]: res = df[df!=0].stack()
In [12]: res
Out[12]:
0 c1 1
c2 1
1 c1 1
2 c2 1
3 c1 1
...
And you can iterate over the original rows:
您可以遍历原始行：
In [13]: res.loc[0]
Out[13]:
c1 1
c2 1
dtype: float64
In [14]: res.loc[0].index
Out[14]: Index(['c1', 'c2'], dtype='object')
Note: I thought you used to be able to return a list in an apply (to create a DataFrame which has list elements) this no longer appears to be the case.
注意：我认为您曾经能够在应用程序中返回一个列表（以创建一个具有列表元素的 DataFrame），但现在似乎不再如此。
