在 Python 中初始化空矩阵

Initialize empty matrix in Python

提问by marriam nayyer

I am trying to convert a MATLAB code in Python. I don't know how to initialize empty matrix in Python.

我正在尝试在 Python 中转换 MATLAB 代码。我不知道如何在 Python 中初始化空矩阵。

MATLAB Code:

MATLAB 代码：

demod4(1) = [];

I tried in Python

我在 Python 中尝试过

demod4[0] = array([])

but it gives error:

但它给出了错误：

only length-1 arrays can be converted to Python scalars

采纳答案by gevang

If you are using numpyarrays, you initialize to 0, by specifying the expected matrix size:

如果您使用的是numpyarrays，则通过指定预期的矩阵大小来初始化为 0：

import numpy as np
d = np.zeros((2,3))

>>> d
    [[ 0.  0.  0.]
     [ 0.  0.  0.]]

This would be the equivalent of MATLAB 's:

这相当于 MATLAB 的：

d = zeros(2,3);

You can also initialize an empty array, again using the expected dimensions/size

您还可以初始化一个空数组，再次使用预期的维度/大小

d = np.empty((2,3))

If you are not using numpy, the closest somewhat equivalent to MATLAB's d = [](i.e., a zero-size matrix) would be using an empty list and then

如果您不使用 numpy，则最接近 MATLAB 的d = []（即零大小矩阵）将使用一个空列表，然后

append values (for filling a vector)

附加值（用于填充向量）

d = []
d.append(0)
d.append(1)
>>> d                                                                     
[0, 1]

or append lists (for filling a matrix row or column):

或附加列表（用于填充矩阵行或列）：

d = []                                                                
d.append(range(0,2))                                                    
d.append(range(2,4))                                                  
>>> d                                                                     
[[0, 1], [2, 3]]

See also:

也可以看看：

initialize a numpy array(SO)

初始化一个 numpy 数组（SO）

NumPy array initialization (fill with identical values)(SO)

NumPy 数组初始化（填充相同的值）（SO）

How do I create an empty array/matrix in NumPy?(SO)

如何在 NumPy 中创建一个空数组/矩阵？（所以）

NumPy for MATLAB users

面向 MATLAB 用户的 NumPy

回答by wflynny

What about initializing a list, populating it, then converting to an array.

初始化一个列表，填充它，然后转换为一个数组怎么样。

demod4 = []  

Or, you could just populate at initialization using a list comprehension

或者，您可以在初始化时使用列表理解进行填充

demod4 = [[func(i, j) for j in range(M)] for i in range(N)]

Or, you could initialize an array of all zeros if you know the size of the array ahead of time.

或者，如果您提前知道数组的大小，则可以初始化一个全为零的数组。

demod4 = [[0 for j in range(M)] for i in range(N)]

or

或者

demod4 = [[0 for i in range(M)]*N]

Or try using numpy.

或者尝试使用numpy.

import numpy as np

N, M = 100, 5000
np.zeros((N, M))

回答by darmat

You could use a nested list comprehension:

您可以使用嵌套列表理解：

# size of matrix n x m
matrix = [ [ 0 for i in range(n) ] for j in range(m) ]

回答by user8154599

M=[]
n=int(input())
m=int(input())
for j in range(n):
   l=[]
   for k in range(m):
       l.append(0)
   M.append(l)
print(M)

This is the traditional way of doing it matrix[m,n], However, python offers many cool ways of doing so as mentioned in other answers.

这是执行 matrix[m,n] 的传统方法，但是，如其他答案中所述，python 提供了许多很酷的方法。

回答by yanefedor

To init matrix with M rows and N columns you can use following pattern:

要使用 M 行 N 列初始化矩阵，您可以使用以下模式：

M = 3
N = 2
matrix = [[0] * N for _ in range(M)]

回答by Luis Mario

rows = 3
columns = 2
M = [[0]*columns]*rows

Or you could also use '' instead of 0

或者你也可以使用 '' 而不是 0

print(M)

Output:

输出：

M = [[0, 0], [0, 0], [0, 0]]