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问题:
I am new to Python 3.4 and I usually use MATLAB/ GNU Octave for matrix calculation. I know we can perform matrix calculation using numpy in Python 2.x, but numpy does not work for Python 3.4.
I tried to create this code to input an m by n matrix. I intended to input [[1,2,3],[4,5,6]]
but the code yields [[4,5,6],[4,5,6]
. Same things happen when I input other m by n matrix, the code yields an m by n matrix whose rows are identical.
Perhaps you can help me to find what is wrong with my code.
m = int(input('number of rows, m = '))
n = int(input('number of columns, n = '))
matrix = []; columns = []
# initialize the number of rows
for i in range(0,m):
matrix += [0]
# initialize the number of columns
for j in range (0,n):
columns += [0]
# initialize the matrix
for i in range (0,m):
matrix[i] = columns
for i in range (0,m):
for j in range (0,n):
print ('entry in row: ',i+1,' column: ',j+1)
matrix[i][j] = int(input())
print (matrix)
回答1:
The problem is on the initialization step.
for i in range (0,m):
matrix[i] = columns
This code actually makes every row of your matrix
refer to the same columns
object. If any item in any column changes - every other column will change:
>>> for i in range (0,m):
... matrix[i] = columns
...
>>> matrix
[[0, 0, 0], [0, 0, 0]]
>>> matrix[1][1] = 2
>>> matrix
[[0, 2, 0], [0, 2, 0]]
You can initialize your matrix in a nested loop, like this:
matrix = []
for i in range(0,m):
matrix.append([])
for j in range(0,n):
matrix[i].append(0)
or, in a one-liner by using list comprehension:
matrix = [[0 for j in range(n)] for i in range(m)]
or:
matrix = [x[:] for x in [[0]*n]*m]
See also:
- How to initialize a two-dimensional array in Python?
Hope that helps.
回答2:
you can accept a 2D list in python this way ...
simply
arr2d = [[j for j in input().strip()] for i in range(n)]
# n is no of rows
for characters
n = int(input().strip())
m = int(input().strip())
a = [[0]*n for _ in range(m)]
for i in range(n):
a[i] = list(input().strip())
print(a)
or
n = int(input().strip())
n = int(input().strip())
a = []
for i in range(n):
a[i].append(list(input().strip()))
print(a)
for numbers
n = int(input().strip())
m = int(input().strip())
a = [[0]*n for _ in range(m)]
for i in range(n):
a[i] = [int(j) for j in input().strip().split(" ")]
print(a)
where n is no of elements in columns while m is no of elements in a row.
In pythonic way, this will create a list of list
回答3:
Apart from the accepted answer, you can also initialise your rows in the following manner -
matrix[i] = [0]*n
Therefore, the following piece of code will work -
m = int(input('number of rows, m = '))
n = int(input('number of columns, n = '))
matrix = []
# initialize the number of rows
for i in range(0,m):
matrix += [0]
# initialize the matrix
for i in range (0,m):
matrix[i] = [0]*n
for i in range (0,m):
for j in range (0,n):
print ('entry in row: ',i+1,' column: ',j+1)
matrix[i][j] = int(input())
print (matrix)
回答4:
This code takes number of row and column from user then takes elements and displays as a matrix.
m = int(input('number of rows, m : '))
n = int(input('number of columns, n : '))
a=[]
for i in range(1,m+1):
b = []
print("{0} Row".format(i))
for j in range(1,n+1):
b.append(int(input("{0} Column: " .format(j))))
a.append(b)
print(a)
回答5:
row=list(map(int,input().split())) #input no. of row and column
b=[]
for i in range(0,row[0]):
a=list(map(int,input().split()))
b.append(a)
print(b)
input
2 3
1 2 3
4 5 6
output
[[1, 2, 3], [4, 5, 6]]
回答6:
If your matrix is given in row manner like below, where size is s*s here s=5
5
31 100 65 12 18 10 13 47 157 6 100 113 174 11 33 88 124 41 20 140 99 32 111 41 20
then you can use this
s=int(input())
b=list(map(int,input().split()))
arr=[[b[j+s*i] for j in range(s)]for i in range(s)]
your matrix will be 'arr'
回答7:
Creating matrix with prepopulated numbers can be done with list comprehension. It may be hard to read but it gets job done:
rows = int(input('Number of rows: '))
cols = int(input('Number of columns: '))
matrix = [[i + cols * j for i in range(1, cols + 1)] for j in range(rows)]
with 2 rows and 3 columns matrix will be [[1, 2, 3], [4, 5, 6]], with 3 rows and 2 columns matrix will be [[1, 2], [3, 4], [5, 6]] etc.
回答8:
a = []
b = []
m=input("enter no of rows: ")
n=input("enter no of coloumns: ")
for i in range(n):
a = []
for j in range(m):
a.append(input())
b.append(a)
Input : 1 2 3 4 5 6 7 8 9
Output : [ ['1', '2', '3'], ['4', '5', '6'], ['7', '8', '9'] ]
回答9:
row=list(map(int,input().split())) #input no. of row and column
b=[]
for i in range(0,row[0]):
print('value of i: ',i)
a=list(map(int,input().split()))
print(a)
b.append(a)
print(b)
print(row)
Output:
2 3
value of i:0
1 2 4 5
[1, 2, 4, 5]
value of i: 1
2 4 5 6
[2, 4, 5, 6]
[[1, 2, 4, 5], [2, 4, 5, 6]]
[2, 3]
Note: this code in case of control.it only control no. Of rows but we can enter any number of column we want i.e row[0]=2
so be careful. This is not the code where you can control no of columns.
回答10:
m,n=map(int,input().split()) # m - number of rows; n - number of columns;
matrix = [[int(j) for j in input().split()[:n]] for i in range(m)]
for i in matrix:print(i)