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将数据填充到矩阵中,传递其定义的索引遇到索引器的元素

matrix numpy python-2.7 arrays python

Jan · 技术社区 · 8 年前

numpy

row = np.array([1,5,8,15,2])
column = np.array([2,3,4,7,8])

mymatrix的维数为10×10

mymatrix = np.zeros(shape=(10,10))

(row=15,column=7)

try:
  mymatrix[row, column] = 100
except IndexError:
  pass

3 回复 | 直到 8 年前

Divakar 8 年前

使用有效行和列的掩码,并使用它来掩码它们,索引到2D数组中,然后用 integer-array indexing row 和 column 作为数组,我们将有一个这样的解决方案-

m,n = mymatrix.shape
mask = (row>=0) & (row < m) & (column>=0) & (column < n)
mymatrix[row[mask], column[mask]] = 100

样本运行-

In [19]: row = np.array([1,5,8,15,2])
    ...: column = np.array([2,3,4,7,8])
    ...: 

In [20]: mymatrix = np.zeros(shape=(10,10))

In [21]: m,n = mymatrix.shape

In [22]: mask = (row>=0) & (row < m) & (column>=0) & (column < n)

In [23]: mymatrix[row[mask], column[mask]] = 100

In [24]: (mymatrix==100).sum() # number of elements edited to 100
Out[24]: 4

一行 和 柱 0 (row>=0) 和 (column>=0) 部分。

glegoux 8 年前

import numpy as np

m = np.zeros((10, 10))
rows = np.array([1, 5, 8, 15, 2])
columns = np.array([2, 3, 4, 7, 8])

n, p = m.shape
for i, j in zip(rows, columns):
    if 0 <= i < n and 0 <= j < p:
        m[i, j] = 100

print(m)

输出:

[[   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.  100.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.  100.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.  100.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.  100.    0.    0.    0.    0.    0.]
 [   0.    0.    0.    0.    0.    0.    0.    0.    0.    0.]]

您可以查看是否正确:

print('rows: {}, \ncolumns: {}'.format(*np.where(m == 100)))

rows: [1 2 5 8], 
columns: [2 8 3 4]

性能:

解决方案.py

import numpy as np

m = np.zeros((10, 10))
rows = np.array([1, 5, 8, 15, 2])
cols = np.array([2, 3, 4, 7, 8]) 

# without numpy
def multiple_insert(m, value, rows, cols):
    n, p = m.shape
    for i, j in zip(rows, cols):
        if 0 <= i < n and 0 <= j < p:
            m[i, j] = value


# with numpy
def multiple_insert2(m, value, rows, cols):
    n, p = m.shape
    mask = (0 <= rows) & (rows < n) & (0 <= cols) & (cols < p)
    m[rows[mask], cols[mask]] = value

ipython 慰问:

In [1]: run solution.py

In [2]: %timeit multiple_insert(m, 100, rows, cols)
100000 loops, best of 3: 4.06 Âµs per loop

In [3]: %timeit multiple_insert2(m, 100, rows, cols)
100000 loops, best of 3: 7.34 Âµs per loop

有时使用always NumPy是不好的。

Saurabh Shrivastava 8 年前

mymatrix[row, column] = 100 . row column 数组不是整数。

您需要迭代行和列数组,并在中传递每个元素 mymatrix