blocks|key|231165|text|您可以使用以下命令从CSR矩阵X中删除行0+<+i+<+X.shape[0]+-+1|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|231166|scipy.sparse.vstack([X[:i,+:],+X[i:,+:]])|code-block|syntax|javascript|231167|您可以分别使用X[1:,+:]或X[:-1,+:]删除第一行或最后一行。一次删除多行可能需要使用您自己的函数。|231168|对于CSR以外的其他格式，这可能不一定有效，因为并非所有格式都支持行切片。|231169|entityMap^0|F|1|K|M|0|0|7|8|G|9|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

You can delete row <code>0 &lt; i &lt; X.shape[0] - 1</code> from a CSR matrix <code>X</code> with

<pre><code>scipy.sparse.vstack([X[:i, :], X[i:, :]])
</code></pre>

You can delete the first or the last row with <code>X[1:, :]</code> or <code>X[:-1, :]</code>, respectively. Deleting multiple rows in one gone will probably require rolling your own function.

For other formats than CSR, this might not necessarily work as not all formats support row slicing.

blocks|key|449184|text|对于CSR来说，这可能是最有效的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|449185|def+delete_row_csr(mat,+i):
++++if+not+isinstance(mat,+scipy.sparse.csr_matrix):
++++++++raise+ValueError("works+only+for+CSR+format+--+use+.tocsr()+first")
++++n+=+mat.indptr[i%2B1]+-+mat.indptr[i]
++++if+n+>+0:
++++++++mat.data[mat.indptr[i]:-n]+=+mat.data[mat.indptr[i%2B1]:]
++++++++mat.data+=+mat.data[:-n]
++++++++mat.indices[mat.indptr[i]:-n]+=+mat.indices[mat.indptr[i%2B1]:]
++++++++mat.indices+=+mat.indices[:-n]
++++mat.indptr[i:-1]+=+mat.indptr[i%2B1:]
++++mat.indptr[i:]+-=+n
++++mat.indptr+=+mat.indptr[:-1]
++++mat._shape+=+(mat._shape[0]-1,+mat._shape[1])|code-block|syntax|javascript|449186|在LIL格式中，它甚至更简单：|449187|def+delete_row_lil(mat,+i):
++++if+not+isinstance(mat,+scipy.sparse.lil_matrix):
++++++++raise+ValueError("works+only+for+LIL+format+--+use+.tolil()+first")
++++mat.rows+=+np.delete(mat.rows,+i)
++++mat.data+=+np.delete(mat.data,+i)
++++mat._shape+=+(mat._shape[0]+-+1,+mat._shape[1])|449188|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

For CSR, this is probably the most efficient way to do it in-place:

<pre><code>def delete_row_csr(mat, i):
 if not isinstance(mat, scipy.sparse.csr_matrix):
 raise ValueError("works only for CSR format -- use .tocsr() first")
 n = mat.indptr[i+1] - mat.indptr[i]
 if n &gt; 0:
 mat.data[mat.indptr[i]:-n] = mat.data[mat.indptr[i+1]:]
 mat.data = mat.data[:-n]
 mat.indices[mat.indptr[i]:-n] = mat.indices[mat.indptr[i+1]:]
 mat.indices = mat.indices[:-n]
 mat.indptr[i:-1] = mat.indptr[i+1:]
 mat.indptr[i:] -= n
 mat.indptr = mat.indptr[:-1]
 mat._shape = (mat._shape[0]-1, mat._shape[1])
</code></pre>

In LIL format it's even simpler:

<pre><code>def delete_row_lil(mat, i):
 if not isinstance(mat, scipy.sparse.lil_matrix):
 raise ValueError("works only for LIL format -- use .tolil() first")
 mat.rows = np.delete(mat.rows, i)
 mat.data = np.delete(mat.data, i)
 mat._shape = (mat._shape[0] - 1, mat._shape[1])
</code></pre>

blocks|key|231219|text|请注意，稀疏矩阵在某种程度上支持花式索引。所以你能做的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|231220|mask+=+np.ones(len(mat),+dtype=bool)
mask[rows_to_delete]+=+False
#+unfortunatly+I+think+boolean+indexing+does+not+work:
w+=+np.flatnonzero(mask)
result+=+s[w,:]|code-block|syntax|javascript|231221|delete方法实际上也不会做其他任何事情。|231222|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Note that sparse matrices support fancy indexing to some degree. So what you can do is this:

<pre><code>mask = np.ones(len(mat), dtype=bool)
mask[rows_to_delete] = False
# unfortunatly I think boolean indexing does not work:
w = np.flatnonzero(mask)
result = s[w,:]
</code></pre>

The delete method doesn't really do anything else either.

blocks|key|231300|text|Pv.的答案是一个好的、可靠的就地解决方案，它需要|type|unstyled|depth|inlineStyleRanges|entityRanges|data|231301|a+=+scipy.sparse.csr_matrix((100,100),+dtype=numpy.int8)
%25timeit+delete_row_csr(a.copy(),+0)
10000+loops,+best+of+3:+80.3+us+per+loop|code-block|syntax|javascript|231302|对于任何数组大小。由于布尔索引适用于稀疏矩阵，至少在scipy+>=+0.14.0中是这样，所以我建议在删除多行时使用它：|offset|length|style|CODE|231303|def+delete_rows_csr(mat,+indices):
++++"""
++++Remove+the+rows+denoted+by+``indices``+form+the+CSR+sparse+matrix+``mat``.
++++"""
++++if+not+isinstance(mat,+scipy.sparse.csr_matrix):
++++++++raise+ValueError("works+only+for+CSR+format+--+use+.tocsr()+first")
++++indices+=+list(indices)
++++mask+=+numpy.ones(mat.shape[0],+dtype=bool)
++++mask[indices]+=+False
++++return+mat[mask]|231304|此解决方案删除单行所需的时间要长得多|231305|%25timeit+delete_rows_csr(a.copy(),+[50])
1000+loops,+best+of+3:+509+us+per+loop|231306|但是对于删除多行更有效，因为执行时间几乎不会随着行数的增加而增加。|231307|%25timeit+delete_rows_csr(a.copy(),+numpy.random.randint(0,+100,+30))
1000+loops,+best+of+3:+523+us+per+loop|231308|entityMap^0|0|0|Q|F|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

Pv.s answer is a good and solid in-place solution that takes

<pre><code>a = scipy.sparse.csr_matrix((100,100), dtype=numpy.int8)
%timeit delete_row_csr(a.copy(), 0)
10000 loops, best of 3: 80.3 us per loop
</code></pre>

for any array size. Since boolean indexing works for sparse matrices, at least in <code>scipy &gt;= 0.14.0</code>, I would suggest to use it whenever multiple rows are to be removed:

<pre><code>def delete_rows_csr(mat, indices):
 """
 Remove the rows denoted by ``indices`` form the CSR sparse matrix ``mat``.
 """
 if not isinstance(mat, scipy.sparse.csr_matrix):
 raise ValueError("works only for CSR format -- use .tocsr() first")
 indices = list(indices)
 mask = numpy.ones(mat.shape[0], dtype=bool)
 mask[indices] = False
 return mat[mask]
</code></pre>

This solution takes significantly longer for a single row removal

<pre><code>%timeit delete_rows_csr(a.copy(), [50])
1000 loops, best of 3: 509 us per loop
</code></pre>

But is more efficient for the removal of multiple rows, as the execution time barely increases with the number of rows

<pre><code>%timeit delete_rows_csr(a.copy(), numpy.random.randint(0, 100, 30))
1000 loops, best of 3: 523 us per loop
</code></pre>

blocks|key|449327|text|要从A中删除第i行，只需使用左矩阵乘法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|449328|B+=+J*A|code-block|syntax|javascript|449329|其中J是去掉第i行的稀疏单位矩阵。|449330|左乘J的转置将插入一个零向量返回到B的第i行，这使得这个解决方案更一般。|449331|A0+=+J.T+*+B|449332|为了构造J本身，我在稀疏对角线矩阵上使用了pv的解决方案，如下所示(对于这种特殊情况，可能有更简单的解决方案？)|449333|def+identity_minus_rows(N,+rows):
++++if+np.isscalar(rows):
++++++++rows+=+[rows]
++++J+=+sps.diags(np.ones(N),+0).tocsr()++#+make+a+diag+matrix
++++for+r+in+sorted(rows):
++++++++J+=+delete_row_csr(J,+r)
++++return+J|449334|也可以通过右乘以适当大小的J.T来删除列。|449335|最后，乘法在这种情况下是有效的，因为J是如此稀疏。|449336|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

To remove the i'th row from A simply use left matrix multiplication:

<pre><code>B = J*A
</code></pre>

where J is a sparse identity matrix with i'th row removed. 
Left multiplication by the transpose of J will insert a zero-vector back to the i'th row of B, which makes this solution a bit more general.

<pre><code>A0 = J.T * B
</code></pre>

To construct J itself, I used pv.'s solution on a sparse diagonal matrix as follows (maybe there's a simpler solution for this special case?)

<pre><code>def identity_minus_rows(N, rows):
 if np.isscalar(rows):
 rows = [rows]
 J = sps.diags(np.ones(N), 0).tocsr() # make a diag matrix
 for r in sorted(rows):
 J = delete_row_csr(J, r)
 return J
</code></pre>

You may also remove columns by right-multiplying by J.T of the appropriate size. 
Finally, multiplication is efficient in this case because J is so sparse.

blocks|key|231383|text|使用@loli实现，我在这里留下一个删除列的函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|231384|def+delete_cols_csr(mat,+indices):
++++"""
++++Remove+the+cols+denoted+by+``indices``+form+the+CSR+sparse+matrix+``mat``.
++++"""
++++if+not+isinstance(mat,+csr_matrix):
++++++++raise+ValueError("works+only+for+CSR+format+--+use+.tocsr()+first")
++++indices+=+list(indices)
++++mask+=+np.ones(mat.shape[1],+dtype=bool)
++++mask[indices]+=+False
++++return+mat[:,mask]|code-block|syntax|javascript|231385|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using @loli implementation, here I leave a function to remove columns:

<pre><code>def delete_cols_csr(mat, indices):
 """
 Remove the cols denoted by ``indices`` form the CSR sparse matrix ``mat``.
 """
 if not isinstance(mat, csr_matrix):
 raise ValueError("works only for CSR format -- use .tocsr() first")
 indices = list(indices)
 mask = np.ones(mat.shape[1], dtype=bool)
 mask[indices] = False
 return mat[:,mask]
</code></pre>

Let's say I have a 2-dimensional matrix as a numpy array. If I want to delete rows with specific indices in this matrix, I use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.delete.html?highlight=delete#numpy.delete" rel="noreferrer"><code>numpy.delete()</code></a>. Here is an example of what I mean:

<pre><code>In [1]: my_matrix = numpy.array([
 ...: [10, 20, 30, 40, 50],
 ...: [15, 25, 35, 45, 55],
 ...: [95, 96, 97, 98, 99]
 ...: ])
In [2]: numpy.delete(my_matrix, [0, 2], axis=0)
Out[2]: array([[15, 25, 35, 45, 55]])
</code></pre>

I'm looking for a way to do the above with matrices from the <a href="http://docs.scipy.org/doc/scipy/reference/sparse.html" rel="noreferrer"><code>scipy.sparse</code></a> package. I know it's possible to do this by converting the entire matrix into a numpy array but I don't want to do that. Is there any other way of doing that? 

Thanks a lot!

Is there a numpy.delete() equivalent for sparse matrices?

假设我有一个二维矩阵作为numpy数组。如果我想删除这个矩阵中具有特定索引的行，我会使用。下面是我的意思的一个例子：In [1]: my_matrix = numpy.array([   ...:     [10, 20, 30, 40, 50],   ...:     [15, 25, 35, 45, 55],   ...

问稀疏矩阵是否存在numpy.delete()等效项？
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