Random size list generation in Python

Question

I want to create a random size list in Python.

Here's my code:

row_num = int(input('Please enter matrix row number:'))
colum_num = int(input('please enter matrix column number:'))
a = [[0]*row_num]*colum_num
print("Please enter matrix: ")
for i in range(colum_num):
    for j in range(row_num):
        a[i][j]=int(input())

print(a)

For example, I want to generate a 2*3 matrix, here's my outcome:

Please enter matrix row number:2
please enter matrix column number:3
Please enter matrix: 
1
2
3
4
5
6
[[5, 6], [5, 6], [5, 6]]

I expect the result to be [[1,2],[3,4],[5,6]].

Why is the result the last element?

full answers already given, but key is try: a = [[0]*3]*3 #n print(a) #n a[1][1] = "Same" #n print(a) #n — Joop
– Joop, Commented Oct 2, 2015 at 7:53
@Joop, the same problem will remain due to the reasons I gave in my answer. Try running it in IDLE. — Ilya Peterov
– Ilya Peterov, Commented Oct 2, 2015 at 8:06
I was not pretending to give answer just succinctly stating problem as you did in your answer — Joop
– Joop, Commented Oct 2, 2015 at 13:30

Ilya Peterov · Accepted Answer · 2015-10-02 07:27:21Z

The problem in your code is that when you do something like that:

a = [[0] * row_num] * colum_num

you create colum_num links to list [[0] * row_num]. When you change one of the lists (in your for), others get changed too. You can run following code to see what I mean

a = [1,2,3]
b = [a] * 3
print(b)
a[1] = 10
print(b)

the output will be [[1, 2, 3], [1, 2, 3], [1, 2, 3]] and [[1, 10, 3], [1, 10, 3], [1, 10, 3]] because b is just 3 links to a.

There are multiple ways of creating new lists and not copies, e.g. indexing the full list

b = [a[:]] * 3

or using copy module

import copy
b = [copy.copy(a)] * 3

or using generator expressions to create multiple lists that are alike (full of 0), but are different objects, like sgbirch suggested.

a = [[0 for x in range(column_num)] for y in range(row_num)]

By the way, there is no pont of creating a list full of 0, you can just do something like that:

row_num = int(input('Please enter matrix row number:'))
column_num = int(input('please enter matrix column number:'))
a = [[int(input()) for x in range(column_num)] for y in range(row_num)]

it also is probably simpler for user to input the matrix row by row.

a = [[int(x) for x in input('Input row № {} ({} numbers): '.format(y + 1, column_num)).split()] for y in range(row_num)]

sgbirch · Accepted Answer · 2017-05-29 07:53:04Z

You have to create a list which contains a list of lists.

# Create a list containing lists initialised to 0
a = [[0 for x in range(row_num)] for x in range(column_num)]

Your code is actually producing a colum_length list whose elements all contain a reference to the same row_num length lists. You can see this in the following code based on yours:

row_num = 2                                                                     
colum_num = 3                                                                   

a = [[0 for x in range(row_num)] for x in range(colum_num)]                     
b = [[0]*row_num]*colum_num                                                     

assert a == b                                                                   

count = 0                                                                       
for i in range(colum_num):                                                      
    for j in range(row_num):                                                    
        a[i][j] = count                                                         
        b[i][j] = count                                                         
        print "a",a                                                             
        print "b",b                                                             
        print                                                                   
        count += 1

Which produces:

a [[0, 0], [0, 0], [0, 0]]
b [[0, 0], [0, 0], [0, 0]]

a [[0, 1], [0, 0], [0, 0]]
b [[0, 1], [0, 1], [0, 1]]

a [[0, 1], [2, 0], [0, 0]]
b [[2, 1], [2, 1], [2, 1]]

a [[0, 1], [2, 3], [0, 0]]
b [[2, 3], [2, 3], [2, 3]]

a [[0, 1], [2, 3], [4, 0]]
b [[4, 3], [4, 3], [4, 3]]

a [[0, 1], [2, 3], [4, 5]]
b [[4, 5], [4, 5], [4, 5]]

An alternative approach might be to use numpy which provides ready made array primitives:

import numpy
# Returns a new array of given shape and type, filled with zeros.
a = numpy.zeros((2, 3), dtype=numpy.int)
print a
[[0 0 0]
 [0 0 0]]

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