Python 根据单个单元格上的条件从numpy数组中删除列
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Delete columns from numpy array depending on a condition on a single cell
提问by user3176697
Above all, sorry for my bad English.
最重要的是，对不起我的英语不好。
I have this array t:
我有这个数组t：
array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9],
[ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
I would like to delete the columns where the value of second line is null. Here, I would like to delete the columns 0, 2, 4, 6 and 9, to obtain this array:
我想删除第二行值为空的列。在这里，我想删除第 0、2、4、6 和 9 列，以获取此数组：
array([[ 1, 0, 5, 7, 8 ],
[ 11, 13, 15, 17, 18 ]])
I tried with np.sum()but didn't succeed.
我尝试过np.sum()但没有成功。
采纳答案by Eelco Hoogendoorn
Similar to Juh_, but more expressive, and avoiding some minor unnecessary performance overhead. A grand total of 12 highly pythonic, explicit and unambigious characters. This is really numpy 101; if you are still trying to wrap your head around this, you would do yourself a favor by reading a numpy primer.
类似于 Juh_，但更具表现力，并避免了一些不必要的小性能开销。总共 12 个高度 pythonic、明确和明确的字符。这真的是 numpy 101；如果您仍在试图解决这个问题，那么您可以通过阅读一本 numpy 入门书来帮自己一个忙。
import numpy as np
a = np.array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9],
[ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
print a[:,a[1]!=0]
回答by Krzysztof Rosiński
I got this working like this:
我得到了这样的工作：
data = array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
res = array([(a, b,) for a, b in zip(data[0], data[1]) if b]).transpose()
got the result
得到了结果
In [23]: res
Out[23]:
array([[ 1, 0, 5, 7, 8],
[11, 13, 15, 17, 18]])
回答by eskaev
With numpy.delete:
a = np.array([[0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
indices = [i for (i,v) in enumerate(a[1]) if v==0]
# [0, 2, 4, 6, 9]
a = np.delete(a, indices, 1)
# array([[ 1, 0, 5, 7, 8], [11, 13, 15, 17, 18]])
回答by Juh_
Simple (fully numpy) solution:
简单（完全麻木）的解决方案：
import numpy as np
t = np.array([[ 0, 1, 2, 0, 4, 5, 6, 7, 8, 9], [ 0, 11, 0, 13, 0, 15, 0, 17, 18, 0]])
indices_to_keep = t[1].nonzero()[0]
print t[:,indices_to_keep]
# [[ 1 0 5 7 8]
# [11 13 15 17 18]]
