blocks|key|1575208|text|通过利用np.accumulate，您可以在没有任何循环的情况下完成此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1575209|R+=+5+#+100000
C+=+3+#+100

mat+++=+np.random.randint(2,+size=(R,C))
print(mat)+#+original+matrix

state++++=+np.zeros((1,C))++++++++++++++++++++++++#+or+np.ones((1,C))
mat++++++=+np.concatenate([state,mat])++++++++++++#+insert+state+row
zRows++++=+np.isin(np.sum(mat,1),[0,C])+++++++++++#+all+zeroes+or+all+ones
iRows++++=+np.arange(R%2B1)+*+zRows.astype(np.int)++#+base+indexes
mat++++++=+mat[np.maximum.accumulate(iRows)][1:]++#+indirection,+remove+state
print(mat)+#+modified

#original
[[0+0+1]
+[1+1+1]
+[1+0+1]
+[0+0+0]
+[1+0+1]]

#+modified
[[0+0+0]
+[1+1+1]
+[1+1+1]
+[0+0+0]
+[0+0+0]]|code-block|syntax|javascript|1575210|它的工作方式是为需要更改的行准备一个间接数组。这是通过行索引的np.arange完成的，其中我们将需要替换的索引设置为零。累积最大索引会将每个被替换的行映射到它之前的全零或全一行。|1575211|例如：|1575212|++[+0,+1,+2,+3,+4,+5+]+#+row+indexes
++[+0,+1,+0,+0,+1,+0+]+#+rows+that+are+all+zeroes+or+all+ones+(zRows)
++[+0,+1,+0,+0,+4,+0+]+#+multiplied+(iRows)
++[+0,+1,+1,+1,+4,+4+]+#+np.maximum.accumulate|1575213|这为我们提供了一个索引列表，其中的行内容应该从中获取。|1575214|状态由在执行操作之前在矩阵开始处插入的额外行表示，在执行操作后删除。|1575215|这个解决方案对于非常小的矩阵(5x3)会稍微慢一些，但对于较大的矩阵(100000x100:+0.7秒对14秒)，它可以让您的速度提高20倍。|1575216|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.ufunc.accumulate.html^0|4|D|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|C|$]]|$1|M|3|N|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|C|$]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

You can do this without any loops by leveraging <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.ufunc.accumulate.html" rel="nofollow noreferrer">np.accumulate</a>:

<pre><code>R = 5 # 100000
C = 3 # 100

mat = np.random.randint(2, size=(R,C))
print(mat) # original matrix

state = np.zeros((1,C)) # or np.ones((1,C))
mat = np.concatenate([state,mat]) # insert state row
zRows = np.isin(np.sum(mat,1),[0,C]) # all zeroes or all ones
iRows = np.arange(R+1) * zRows.astype(np.int) # base indexes
mat = mat[np.maximum.accumulate(iRows)][1:] # indirection, remove state
print(mat) # modified

#original
[[0 0 1]
 [1 1 1]
 [1 0 1]
 [0 0 0]
 [1 0 1]]

# modified
[[0 0 0]
 [1 1 1]
 [1 1 1]
 [0 0 0]
 [0 0 0]]
</code></pre>

The way it works is by preparing an indirection array for rows that need to be changed. This is done from an np.arange of row indexes in which we set to zero the the indexes that will need replacement. Accumulating the maximum index will map each replaced row to an all-zero or all-one row before it.

For example:

<pre><code> [ 0, 1, 2, 3, 4, 5 ] # row indexes
 [ 0, 1, 0, 0, 1, 0 ] # rows that are all zeroes or all ones (zRows)
 [ 0, 1, 0, 0, 4, 0 ] # multiplied (iRows)
 [ 0, 1, 1, 1, 4, 4 ] # np.maximum.accumulate
</code></pre>

This gives us a list of indexes where row content should be taken from.

The state is represented by an extra row inserted at the beginning of the matrix before performing the operation and removed afterward.

This solution will be marginally slower for very small matrices (5x3) but it can give you a 20x speed boost for larger ones (100000x100: 0.7 second vs 14 seconds).

blocks|key|1790496|text|下面是一个简单快速的numpy方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1790497|import+numpy+as+np

def+pp():
++++m,n+=+a.shape
++++A+=+a.sum(axis=1)++++
++++A+=+np.where((A==0)%7C(A==n))[0]
++++if+not+A.size:
++++++++return+np.ones_like(a)+if+state+else+np.zeros_like(a)
++++st+=+np.concatenate([np.arange(A[0]!=0),+A,+[m]])
++++v+=+a[st[:-1],0]
++++if+A[0]:
++++++++v[0]+=+state
++++return+np.broadcast_to(v.repeat(st[1:]-st[:-1])[:,None],(m,n))|code-block|syntax|javascript|1790498|我用这个做了一些计时|1790499|state=0
a+=+(np.random.random((100000,100))<np.random.random((100000,1))).astype(int)|1790500|简单的测试用例：|1790501|0.8655898020006134+++#+me
4.089095343002555++++#+Alain+T.
2.2958932030014694+++#+Divakar+1
2.2178015549980046+++#+&+2|1790502|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Here is a simple and fast numpy method:

<pre><code>import numpy as np

def pp():
 m,n = a.shape
 A = a.sum(axis=1) 
 A = np.where((A==0)|(A==n))[0]
 if not A.size:
 return np.ones_like(a) if state else np.zeros_like(a)
 st = np.concatenate([np.arange(A[0]!=0), A, [m]])
 v = a[st[:-1],0]
 if A[0]:
 v[0] = state
 return np.broadcast_to(v.repeat(st[1:]-st[:-1])[:,None],(m,n))
</code></pre>

I made some timings using this 

<pre><code>state=0
a = (np.random.random((100000,100))&lt;np.random.random((100000,1))).astype(int)
</code></pre>

simple test case:

<pre><code>0.8655898020006134 # me
4.089095343002555 # Alain T.
2.2958932030014694 # Divakar 1
2.2178015549980046 # &amp; 2
</code></pre>

Suppose I have a matrix that is 100000 x 100

<pre><code>import numpy as np

mat = np.random.randint(2, size=(100000,100))
</code></pre>

I wish to go through this matrix, and if each <code>row</code> contains entirely either 1 or 0 I wish to change a <code>state</code> variable to that value. If the state is not changed, I wish to set the entire <code>row</code> the value of <code>state</code>. The initial value of <code>state</code> is 0. 

Naively in a <code>for</code> loop this can be done as follows

<pre><code>state = 0

for row in mat:
 if set(row) == {1}:
 state = 1
 elif set(row) == {0}:
 state = 0
 else:
 row[:] = state
</code></pre>

However, when the size of the matrix increases this takes an impractical amount of time. Could someone point me in the direction in how to leverage <code>numpy</code> to vectorize this loop and speed it up?

So for a sample input

<pre><code>array([[0, 1, 0],
 [0, 0, 1],
 [1, 1, 1],
 [0, 0, 1],
 [0, 0, 1]])
</code></pre>

The expected output in this case would be

<pre><code>array([[0, 0, 0],
 [0, 0, 0],
 [1, 1, 1],
 [1, 1, 1],
 [1, 1, 1]])
</code></pre>

How to vectorize a loop through a matrix numpy

import numpy as npmat = np.random.randint(2, size=(100000,100))我希望遍历这个矩阵，如果每个row完全包含1或0，我希望将state变量更改为该值。state的初始值为0。state = 0for row in mat:    if set(row) == ...

问如何通过矩阵numpy向量化循环
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