I have to print this python code in a 5x5 array the array should look like this :
0 1 4 (infinity) 3
1 0 2 (infinity) 4
4 2 0 1 5
(inf)(inf) 1 0 3
3 4 5 3 0
can anyone help me print this table? using indices.
for k in range(n):
for i in range(n):
for j in range(n):
if A[i][k]+A[k][j]<A[i][j]:
A[i][j]=A[i][k]+A[k][j]
This question is related to
python
I used numpy to generate the array, but list of lists array should work similarly.
import numpy as np
def printArray(args):
print "\t".join(args)
n = 10
Array = np.zeros(shape=(n,n)).astype('int')
for row in Array:
printArray([str(x) for x in row])
If you want to only print certain indices:
import numpy as np
def printArray(args):
print "\t".join(args)
n = 10
Array = np.zeros(shape=(n,n)).astype('int')
i_indices = [1,2,3]
j_indices = [2,3,4]
for i in i_indices:printArray([str(Array[i][j]) for j in j_indices])
There is always the easy way.
import numpy as np
print(np.matrix(A))
for i in A:
print('\t'.join(map(str, i)))
In addition to the simple print answer, you can actually customise the print output through the use of the numpy.set_printoptions function.
Prerequisites:
>>> import numpy as np
>>> inf = np.float('inf')
>>> A = np.array([[0,1,4,inf,3],[1,0,2,inf,4],[4,2,0,1,5],[inf,inf,1,0,3],[3,4,5,3,0]])
The following option:
>>> np.set_printoptions(infstr="(infinity)")
Results in:
>>> print(A)
[[ 0. 1. 4. (infinity) 3.]
[ 1. 0. 2. (infinity) 4.]
[ 4. 2. 0. 1. 5.]
[(infinity) (infinity) 1. 0. 3.]
[ 3. 4. 5. 3. 0.]]
The following option:
>>> np.set_printoptions(formatter={'float': "\t{: 0.0f}\t".format})
Results in:
>>> print(A)
[[ 0 1 4 inf 3 ]
[ 1 0 2 inf 4 ]
[ 4 2 0 1 5 ]
[ inf inf 1 0 3 ]
[ 3 4 5 3 0 ]]
If you just want to have a specific string output for a specific array, the function numpy.array2string is also available.
print(mat.__str__())
where mat is variable refering to your matrix object
A combination of list comprehensions and str
joins can do the job:
inf = float('inf')
A = [[0,1,4,inf,3],
[1,0,2,inf,4],
[4,2,0,1,5],
[inf,inf,1,0,3],
[3,4,5,3,0]]
print('\n'.join([''.join(['{:4}'.format(item) for item in row])
for row in A]))
yields
0 1 4 inf 3
1 0 2 inf 4
4 2 0 1 5
inf inf 1 0 3
3 4 5 3 0
Using for-loops with indices is usually avoidable in Python, and is not considered "Pythonic" because it is less readable than its Pythonic cousin (see below). However, you could do this:
for i in range(n):
for j in range(n):
print '{:4}'.format(A[i][j]),
print
The more Pythonic cousin would be:
for row in A:
for val in row:
print '{:4}'.format(val),
print
However, this uses 30 print statements, whereas my original answer uses just one.
using indices, for loops and formatting:
import numpy as np
def printMatrix(a):
print "Matrix["+("%d" %a.shape[0])+"]["+("%d" %a.shape[1])+"]"
rows = a.shape[0]
cols = a.shape[1]
for i in range(0,rows):
for j in range(0,cols):
print "%6.f" %a[i,j],
print
print
def printMatrixE(a):
print "Matrix["+("%d" %a.shape[0])+"]["+("%d" %a.shape[1])+"]"
rows = a.shape[0]
cols = a.shape[1]
for i in range(0,rows):
for j in range(0,cols):
print("%6.3f" %a[i,j]),
print
print
inf = float('inf')
A = np.array( [[0,1.,4.,inf,3],
[1,0,2,inf,4],
[4,2,0,1,5],
[inf,inf,1,0,3],
[3,4,5,3,0]])
printMatrix(A)
printMatrixE(A)
which yields the output:
Matrix[5][5]
0 1 4 inf 3
1 0 2 inf 4
4 2 0 1 5
inf inf 1 0 3
3 4 5 3 0
Matrix[5][5]
0.000 1.000 4.000 inf 3.000
1.000 0.000 2.000 inf 4.000
4.000 2.000 0.000 1.000 5.000
inf inf 1.000 0.000 3.000
3.000 4.000 5.000 3.000 0.000
Source: Stackoverflow.com