在矩阵中将数字更改为零

I have the following matrix:

我有以下矩阵:

a = array([
   [100. , 100., 100.],
   [175.2, 198.,  32.],
   [ 38. , 82. ,  38.],
   [155. , 32. ,  23.],
   [ 38. , 67. ,  30.]])

How do I change the numbers in all rows, but except the last row into zero if a number of a row is not equal to 38 and 32? What I want to get is as follows:

如何更改所有行中的数字,但如果一行数不等于38和32,则除最后一行外的数字为零?我想得到的是如下:

a = array([
   [ 0 ,  0 ,  0 ],
   [ 0 ,  0 , 32.],
   [38.,  0 , 38.],
   [ 0 , 32.,  0 ],
   [38., 67., 30.]])

And then I want to leave just the earliest number that appears in each column. Something like this:

然后我想留下每列中出现的最早的数字。像这样的东西:

a = array([
   [ 0 ,  0 ,  0 ],
   [ 0 ,  0 , 32.],
   [38.,  0 ,  0 ],
   [ 0 , 32.,  0 ],
   [ 0 ,  0 ,  0 ]])

3 个解决方案

#1

This should do it:

这应该这样做:

positions = [(y, x) for x, y in enumerate(np.argmax(np.isin(a, [32,38]), axis=0))]

result = np.zeros(a.shape)

for p in positions:
    result[p] = a[p]

#[[ 0 ,  0 ,  0 ],
# [ 0 ,  0 , 32.],
# [38.,  0 ,  0 ],
# [ 0 , 32.,  0 ],
# [ 0 ,  0 ,  0 ]]

#2

Here is a vectorised and in place solution.

这是一个矢量化和就地解决方案。

import numpy as np

a = np.array([[100. , 100., 100.],
              [175.2, 198.,  32.],
              [ 38. , 82. ,  38.],
              [155. , 32. ,  23.],
              [ 38. , 67. ,  30.]])

a[:-1][(a[:-1]!=38) & (a[:-1]!=32)] = 0

# array([[  0.,   0.,   0.],
#        [  0.,   0.,  32.],
#        [ 38.,   0.,  38.],
#        [  0.,  32.,   0.],
#        [ 38.,  67.,  30.]])

You can also use numpy for your final result:

您也可以使用numpy作为最终结果:

idx = ((a!=0).argmax(axis=0), np.array(range(a.shape[1])))
result = np.empty(a.shape)
result[idx] = a[idx]

# array([[  0.,   0.,   0.],
#        [  0.,   0.,  32.],
#        [ 38.,   0.,   0.],
#        [  0.,  32.,   0.],
#        [  0.,   0.,   0.]])

#3

And one more (albeit inefficient) solution:

还有一个(尽管效率低下)解决方案:

import numpy as np

a = np.array([
   [100. , 100., 100.],
   [175.2, 198.,  32.],
   [ 38. , 82. ,  38.],
   [155. , 32. ,  23.],
   [ 38. , 67. ,  30.]])


def what(y,lastRow):
    found = False
    for o in y:
        if o in [32,38] and not found and not lastRow:
            found = True
            yield o
        else:
            yield 0

# probably inefficient due to list comp instead of np-own ops
b = np.array([list(what(y,idy==len(a)-1)) for idy,y in enumerate(a)]) 

print(b)

Output:

[[  0.   0.   0.]
 [  0.   0.  32.]
 [ 38.   0.   0.]
 [  0.  32.   0.]
 [  0.   0.   0.]]

#1