如何在Python中有效地重复矩阵中的二进制模式和比率？

发布于 2025-01-23 18:50:48 字数 194 浏览 1 评论 0原文

我想在python中有效打印一个矩阵，该矩阵遵循5 0s的列中的特定模式，然后在3 1s，然后在5 0s等上等等，依此类推，如下所示，如下所示：1000行：

原文

I want to efficiently print a matrix in python that follows a specific pattern in the columns of 5 0s then 3 1s then 5 0s and so on and so forth as shown below for 1000 rows:

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绿阴红影里的.如风往事 2025-01-30 18:50:48

您可以使用np.block的组合和列表理解：

out = np.block([[np.zeros((5, 4))],[np.ones((3, 4))]] * 125).astype(int)

可以将其功能化以回答类似的问题类似的问题：

def block_repeat(size, *args):
    block = np.block([[a] for a in args])
    return np.resize(block, (size,) + block.shape[1:])

block_repeat(1000, np.zeros((5,4)), np.ones((3,4)))
Out[]: 
array([[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       ...,
       [1., 1., 1., 1.],
       [1., 1., 1., 1.],
       [1., 1., 1., 1.]])

You can use a combination of np.block and list comprehension:

out = np.block([[np.zeros((5, 4))],[np.ones((3, 4))]] * 125).astype(int)

This can be functionalized to answer similar questions like so:

def block_repeat(size, *args):
    block = np.block([[a] for a in args])
    return np.resize(block, (size,) + block.shape[1:])

block_repeat(1000, np.zeros((5,4)), np.ones((3,4)))
Out[]: 
array([[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.],
       ...,
       [1., 1., 1., 1.],
       [1., 1., 1., 1.],
       [1., 1., 1., 1.]])

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生来就爱笑 2025-01-30 18:50:48

您可以做到这一点：

import numpy as np

my_array = np.array([[[0,0,0,0], [0,0,0,0], [0,0,0,0], [0,0,0,0], [0,0,0,0], [1,1,1,1], [1,1,1,1], [1,1,1,1]] for i in range(125)])

您可以检查形状。它有125行8行IE 1000：

>>> my_array.shape
(125, 8, 4)

要打印它可以使用：

count_row = 0
for row in my_array:
    for row2 in row:
        print(row2)
        count_row += 1

输出：

# count_row is equal to 1000.

[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[0 0 0 0]
....

Here's how you can do it:

import numpy as np

my_array = np.array([[[0,0,0,0], [0,0,0,0], [0,0,0,0], [0,0,0,0], [0,0,0,0], [1,1,1,1], [1,1,1,1], [1,1,1,1]] for i in range(125)])

You can check the shape. It has 125 rows of 8 rows i.e. 1000:

>>> my_array.shape
(125, 8, 4)

To print it you can use:

count_row = 0
for row in my_array:
    for row2 in row:
        print(row2)
        count_row += 1

Output:

# count_row is equal to 1000.

[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[0 0 0 0]
....

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童话 2025-01-30 18:50:48

这没有使用numpy，但它可以完成工作。您以后可以使用numpy.matrix将其转换为numpy矩阵。

import itertools

cycle = itertools.cycle([("0") for i in range(5)] + [("1") for i in range(3)])

for i in range(1000):
  item = next(cycle)
  print(4 * item)

输出 -

This didn't use numpy but it does the job. You can later convert it into a numpy matrix using numpy.matrix.

import itertools

cycle = itertools.cycle([("0") for i in range(5)] + [("1") for i in range(3)])

for i in range(1000):
  item = next(cycle)
  print(4 * item)

Output -

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辞旧 2025-01-30 18:50:48

本着@ishan高尔夫的精神，这里是一行，没有图书馆：

print("\n".join(["00000111"[i % 8] * 4 for i in range(1000)]))

# Explanation
pattern = "00000111"
for i in range(1000):
    index = i % len(pattern)
    print(pattern[index] * 4)

In the spirit of @Ishan golfing this, here it is in one line, no librarys:

print("\n".join(["00000111"[i % 8] * 4 for i in range(1000)]))

# Explanation
pattern = "00000111"
for i in range(1000):
    index = i % len(pattern)
    print(pattern[index] * 4)

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小梨窩很甜 2025-01-30 18:50:48

使用np.tile，您甚至可以避免使用列表理解：

>>> unit = np.repeat([0] * 5 + [1] * 3, 4)
>>> # you can also use unit = [0] * 20 + [1] * 12,
>>> # but using ndarray as the parameter of np.tile is faster than list.
>>> np.tile(unit, 2).reshape(-1, 4)
array([[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]])
>>> np.tile(unit, 125).reshape(-1, 4)
array([[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       ...,
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]])

Use np.tile, you can even avoid using list comprehension:

>>> unit = np.repeat([0] * 5 + [1] * 3, 4)
>>> # you can also use unit = [0] * 20 + [1] * 12,
>>> # but using ndarray as the parameter of np.tile is faster than list.
>>> np.tile(unit, 2).reshape(-1, 4)
array([[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]])
>>> np.tile(unit, 125).reshape(-1, 4)
array([[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       ...,
       [1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]])

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