Question

# timestable.py

t = int(input("What size of multiplication table would you like to see (enter 1 – 12)?"))
if (t <=12):
    s = input("What kind table would you like to display, enter either R, U, or L? (R for regular full matrix, U for upper triangular matrix, L for lower triangular matrix)?")
    if s=="R":
           for row in range (1,t+1):
              for col in range (1,t+1):
                  prod = row * col
                  if prod < 10:
                      print(' ', end = '')

                  print(row * col, ' ', end = '')

              print()
    elif  s== "L" :


        for i in range(1,t+1):
            for j in range (1, i+1):
                prod = i *j
                if prod < 10:
                  print(' ', end = '')
                print(i * j, ' ', end = '')
            print()
    elif  s== "U" :


        for i in range(1,t+1):
            for j in range (1, i-1):
                prod = i *j
                if prod < 10:
                  print(' ', end = '')
                print(j * i, ' ', end = '')
            print()           



else :
    t= int(input("Please enter a valid multiplecation table size (enter 1 – 12)?"))

Answer 1

如果你想要一个上（下）三角矩阵，你知道位置i，j处的元素x，其中i> j（i

# upper example
t = 6
for i in range(1, t+1):
    for j in range (1, t+1):
        if i > j:
            print("0\t", end=' ') # use whitespace if you want instead of 0
        else:
            prod = i*j
            print("%s\t" % prod, end=' ')
    print()

输出

1    2   3   4   5   6
0    4   6   8   10  12
0    0   9   12  15  18
0    0   0   16  20  24
0    0   0   0   25  30
0    0   0   0   0   36

Answer 2

如果您不介意使用numpy，整个数组处理将更加轻松和紧凑，因此您可以专注于漂亮的打印。

import numpy as np
t = int(input("What size of multiplication table would you like to see (enter 1 – 12)?"))
if (t <=12):
    s = input("What kind table would you like to display?\n" 
              "Enter either R, U, or L? (R for regular full, "
              "U for upper triangular, and L for lower triangular arrays)")

    a = np.arange(1,t+1)
    prod = a.reshape(t,1).dot(a.reshape(1,t))

    if s== "L" :
        prod = np.tril(prod)
    elif s== "U" :
        prod = np.triu(prod)

    for x in range(prod.shape[0]):
        print(*prod[x].tolist(), sep='\t',end=' \n')

else :
    t= int(input("Please enter a valid multiplecation table size (enter 1 – 12)?"))

我试图在矩阵中得到一个上三角形，但我没有得到预期的输出。大家可以帮帮我吗

2 个答案: