如何構造一個numpy陣列來顯示方形方案

Question

我想問是否可以用numpy創建這個方案：

  1 2 3
1       1
2       2
3       3
  1 2 3

...然后使用命令，我將在這些數字的幫助下將“0”或“*”添加到空列，但需要在角落有空列。 如果可能我應該如何編寫代碼？

Answer 1

import numpy as np
n = 3 #number of elements (3 in your example)
a = np.array([[0] * (n + 2)] * (n + 2)) #creating zero-fill 2-d array
for i in range(len(a)):
    for j in range(len(a[i])):
        if i == 0 or i == len(a) - 1: #if we are on a border of array
            if j != 0 and j != len(a[i]) - 1: #but not in corners
                a[i][j] = j
        if j == 0 or j == len(a[i]) - 1:# if we are on a border of array
            if i != 0 and i != len(a) - 1: #but not in corners
                a[i][j] = i
print(a)

Answer 2

要以有效的方式構造數組，使用數組索引和無循環，您可以使用：

import numpy as np

a = np.full((5, 5), ' ')
v = np.array((' ', '1', '2', '3', ' '))

a[0, :] = v
a[-1, :] = v
a[:, 0] = v
a[:, -1] = v

打印 output：

>>> for i in a:
>>>     print(' '.join(i))

  1 2 3  
1       1
2       2
3       3
  1 2 3  

Answer 3

改編@LightVillet 的答案不那么復雜：

import numpy as np
n = 3 #number of elements (3 in your example)
a = np.full((n+2, n+2), ' ') #creating ' '-fill 2-d array

for i in range(1,n+1):
    a[0,i] = i    #top edge
    a[n+2,i] = i  #bottom edge
    a[i,0] = i    #left edge
    a[i,n+2] = i  #right edge

print(a)

復雜度： O(n)

Answer 4

更復雜和低效的方式，

import numpy as np
size = int(input("Please enter size of the Matrix : "))
empty = np.zeros([size,size])
numberCounter = 0
rowNumberCounter = 0
for rowCounter in range(size):
    empty[rowCounter][size-1] = 0
    for colCounter in range(size):
        if rowCounter == 0 and colCounter == 0:
            empty[rowCounter][colCounter] = 0
        if rowCounter == 0 and colCounter < size-1:
            empty[rowCounter][colCounter] = numberCounter
            numberCounter = numberCounter + 1
        if rowCounter == size-1 and colCounter != 0 and colCounter <size-1:
            numberCounter = numberCounter + 1
            empty[rowCounter][colCounter] = numberCounter

    if rowCounter !=0 and rowCounter < size-1:
        empty[rowCounter][0] = rowNumberCounter + 1
        empty[rowCounter][size-1] = rowNumberCounter + 1
        rowNumberCounter = rowNumberCounter + 1
    numberCounter = 0

print(empty)

pushButton = input('Push * to convert 0 to * :')
if pushButton == '*':
        empty_str = empty.astype(str)
        for rowCounter in range(size):
            for colCounter in range(size):
                if empty_str[rowCounter][colCounter] == '0.0':
                    empty_str[rowCounter][colCounter] = '*'

print(empty_str)

Output；

Please enter size of the Matrix : 5
[[0. 1. 2. 3. 0.]
 [1. 0. 0. 0. 1.]
 [2. 0. 0. 0. 2.]
 [3. 0. 0. 0. 3.]
 [0. 1. 2. 3. 0.]]
Push * to convert 0 to * :*
[['*' '1.0' '2.0' '3.0' '*']
 ['1.0' '*' '*' '*' '1.0']
 ['2.0' '*' '*' '*' '2.0']
 ['3.0' '*' '*' '*' '3.0']
 ['*' '1.0' '2.0' '3.0' '*']]