I have a list of numbers of length 10, say
比方说，我有一个长度为10的数字列表
L = [1,2,3,4,5,6,7,8,9,0]
Then, I have an matrix of size 35 * 10 which:
然后，我有一个大小为35 * 10的矩阵，其中：
-
L[0](value 1) can be found in the first column of the matrix - L [0]（值1）可以在矩阵的第一列中找到
-
L[1](value 2) can be found in the 2nd column of the matrix - L [1]（值2）可以在矩阵的第2列中找到
- etc.
- 等等
I would like the index for each value in L in the corresponding column in the matrix given.
我想在给定矩阵的相应列中的L中的每个值的索引。
I know how to do this for one value, but if say the list L is very long and the matrix is large, it might be slow, so wondering if there is a faster way of doing this.
我知道如何为一个值执行此操作，但是如果说列表L很长并且矩阵很大，则可能很慢，因此想知道是否有更快的方法来执行此操作。
1 个解决方案
#1
1
See docs page.
请参阅文档页面。
L = [1,2,3,4,5,6,7,8,9,0]
a = np.random.ranint(0,50,(35,10))
# identify location of value in array that are in L
ix = np.in1d(a.ravel(), L).reshape(a.shape)
Now you just need to determine what to do with your newly-identified indices!
现在您只需要确定如何处理新识别的索引！
By that, I mean you can remove the "bad" values a[ix==False] = np.NAN
这样，我的意思是你可以删除“坏”值a [ix == False] = np.NAN
or operate on the "good" values exclusively a[ix] *= 25
或者仅以[ix] * = 25对“好”值进行操作
#1
1
See docs page.
请参阅文档页面。
L = [1,2,3,4,5,6,7,8,9,0]
a = np.random.ranint(0,50,(35,10))
# identify location of value in array that are in L
ix = np.in1d(a.ravel(), L).reshape(a.shape)
Now you just need to determine what to do with your newly-identified indices!
现在您只需要确定如何处理新识别的索引！
By that, I mean you can remove the "bad" values a[ix==False] = np.NAN
这样，我的意思是你可以删除“坏”值a [ix == False] = np.NAN
or operate on the "good" values exclusively a[ix] *= 25
或者仅以[ix] * = 25对“好”值进行操作