Python Pandas 计算每列中小于 x 的元素数
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Pandas count number of elements in each column less than x
提问by marillion
I have a DataFrame which looks like below. I am trying to count the number of elements less than 2.0 in each column, then I will visualize the result in a bar plot. I did it using lists and loops, but I wonder if there is a "Pandas way" to do this quickly. Thanks!
我有一个如下所示的 DataFrame。我试图计算每列中小于 2.0 的元素数，然后我将在条形图中可视化结果。我使用列表和循环来完成它，但我想知道是否有一种“熊猫方式”可以快速做到这一点。谢谢！
x = []
for i in range(6):
x.append(df[df.ix[:,i]<2.0].count()[i])
then I can get a bar plot using list x.
然后我可以使用 list 获得条形图x。
A B C D E F
0 2.142 1.929 1.674 1.547 3.395 2.382
1 2.077 1.871 1.614 1.491 3.110 2.288
2 2.098 1.889 1.610 1.487 3.020 2.262
3 1.990 1.760 1.479 1.366 2.496 2.128
4 1.935 1.765 1.656 1.530 2.786 2.433
采纳答案by EdChum
In [96]:
df = pd.DataFrame({'a':randn(10), 'b':randn(10), 'c':randn(10)})
df
Out[96]:
a b c
0 -0.849903 0.944912 1.285790
1 -1.038706 1.445381 0.251002
2 0.683135 -0.539052 -0.622439
3 -1.224699 -0.358541 1.361618
4 -0.087021 0.041524 0.151286
5 -0.114031 -0.201018 -0.030050
6 0.001891 1.601687 -0.040442
7 0.024954 -1.839793 0.917328
8 -1.480281 0.079342 -0.405370
9 0.167295 -1.723555 -0.033937
[10 rows x 3 columns]
In [97]:
df[df > 1.0].count()
Out[97]:
a 0
b 2
c 2
dtype: int64
So in your case:
所以在你的情况下：
df[df < 2.0 ].count()
should work
应该管用
EDIT
编辑
some timings
一些时间
In [3]:
%timeit df[df < 1.0 ].count()
%timeit (df < 1.0).sum()
%timeit (df < 1.0).apply(np.count_nonzero)
1000 loops, best of 3: 1.47 ms per loop
1000 loops, best of 3: 560 us per loop
1000 loops, best of 3: 529 us per loop
So @DSM's suggestions are correct and much faster than my suggestion
所以@DSM 的建议是正确的，而且比我的建议快得多
