使用输入在python 3中制作一个没有numpy的矩阵

问题描述 投票:1回答:3

我想要两个输入:a,b或x,y无论什么......当用户输入时说,

3 5

然后shell应该打印一个包含3行和5列的矩阵,它也应该用自然数填充矩阵(数字序列从1开始而不是0)。例::

IN:2 2

OUT:[1,2] [3,4]

python-3.x matrix input range sequence
3个回答
0
投票

如果您的目标只是以该格式获得输出

n,m=map(int,input().split())
count=0
for _ in range(0,n):
    list=[]
    while len(list) > 0 : list.pop()
    for i in range(count,count+m):
        list.append(i)
        count+=1
    print(list)

0
投票

我将尝试不使用numpy库。

row= int(input("Enter number of rows"))
col= int(input("Enter number of columns"))
count= 1
final_matrix= []
for i in range(row):
    sub_matrix= []
    for j in range(col):
        sub_matrix.append(count)
        count += 1
    final_matrix.append(sub_matrix)

0
投票

Numpy库提供了reshape()函数,可以完全满足您的需求。

from numpy import * #import numpy, you can install it with pip
    n = int(input("Enter number of rows: ")) #Ask for your input, edit this as desired.
    m = int(input("Enter number of columns: "))
    x = range(1, n*m+1) #You want the range to start from 1, so you pass that as first argument.
    x = reshape(x,(n,m)) #call reshape function from numpy
    print(x) #finally show it on screen

编辑

如果你不想像评论中指出的那样使用numpy,这里是另一种解决问题的方法,没有任何库。

n = int(input("Enter number of rows: ")) #Ask for your input, edit this as desired.
m = int(input("Enter number of columns: "))
x = 1 #You want the range to start from 1
list_of_lists = [] #create a lists to store your columns
for row in range(n):
    inner_list = []   #create the column
    for col in range(m):
        inner_list.append(x) #add the x element and increase its value
        x=x+1
    list_of_lists.append(inner_list) #add it

for internalList in list_of_lists: #this is just formatting.
    print(str(internalList)+"\n")
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