通过Pandas DataFrame计数每行零个数字? [英] Counting number of zeros per row by Pandas DataFrame?

问题描述

给出一个DataFrame，我想计算每行的零个数.如何使用Pandas进行计算?

Given a DataFrame I would like to compute number of zeros per each row. How can I compute it with Pandas?

这是我目前所做的，返回的索引为零

This is presently what I ve done, this returns indices of zeros

def is_blank(x):
    return x == 0 

indexer = train_df.applymap(is_blank)

推荐答案

使用布尔比较将产生布尔df，然后可以将其强制转换为int，True变为1，False变为0，然后调用count并传递参数axis=1进行逐行计数:

Use a boolean comparison which will produce a boolean df, we can then cast this to int, True becomes 1, False becomes 0 and then call count and pass param axis=1 to count row-wise:

In [56]:

df = pd.DataFrame({'a':[1,0,0,1,3], 'b':[0,0,1,0,1], 'c':[0,0,0,0,0]})
df
Out[56]:
   a  b  c
0  1  0  0
1  0  0  0
2  0  1  0
3  1  0  0
4  3  1  0
In [64]:

(df == 0).astype(int).sum(axis=1)
Out[64]:
0    2
1    3
2    2
3    2
4    1
dtype: int64

打破以上:

In [65]:

(df == 0)
Out[65]:
       a      b     c
0  False   True  True
1   True   True  True
2   True  False  True
3  False   True  True
4  False  False  True
In [66]:

(df == 0).astype(int)
Out[66]:
   a  b  c
0  0  1  1
1  1  1  1
2  1  0  1
3  0  1  1
4  0  0  1

编辑

不需要astype到int，因为在调用sum时Boolean类型将被转换为int，因此简化为:

as pointed out by david the astype to int is unnecessary as the Boolean types will be upcasted to int when calling sum so this simplifies to:

(df == 0).sum(axis=1)

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