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Want to create a matrix like below in python

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

but not sure how to do, i tried list comprehension like below

[[y+x for y in range(4)] for x in range(4)]

which gives output like below

[[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6]]

and also want print the row with maximum sum and columns with maximum sum.

Thanks in advance

Chethu
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4 Answers4

4
[[x * 4 + y + 1 for y in range(4)] for x in range(4)]

which is equivalent to:

[[(x << 2) + y for y in range(1, 5)] for x in range(4)]

Here's a little benchmark:

import timeit


def f1():
    return [[x * 4 + y + 1 for y in range(4)] for x in range(4)]


def f2():
    return [[(x << 2) + y for y in range(1, 5)] for x in range(4)]


def f3():
    a = range(1, 5)
    return [[(x << 2) + y for y in a] for x in range(4)]

N = 5000000
print timeit.timeit('f1()', setup='from __main__ import f1', number=N)
print timeit.timeit('f2()', setup='from __main__ import f2', number=N)
print timeit.timeit('f3()', setup='from __main__ import f3', number=N)

# 13.683984791
# 13.4605276559
# 9.65608339037
# [Finished in 36.9s]

Where we can conclude both f1 & f2 methods give almost the same performance. So it'd be a good choice to calculate the inner range(1,5) only once like f3

BPL
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The range function accepts 3 arguments start, end and step. You can use one range with step 4 and another one for creating the nested lists using the outer range.

>>> [[i for i in range(i, i+4)] for i in range(1, 17, 4)]
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

here is another way with one range:

>>> main_range = range(1, 5)
>>> [[i*j + (i-1)*(4-j) for j in main_range] for i in main_range]
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

And here is a Numpythonic approach:

>>> n = 4
>>> np.split(np.arange(1, n*n + 1), np.arange(n ,n*n, n))
[array([1, 2, 3, 4]), array([5, 6, 7, 8]), array([ 9, 10, 11, 12]), array([13, 14, 15, 16])]
Mazdak
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1

try this,

In [1]: [range(1,17)[n:n+4] for n in range(0, len(range(1,17)), 4)]
Out[1]: [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
Rahul K P
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0

This might be a quick fix

[([x-3,x-2,x-1,x]) for x in range(1,17) if x%4 ==0]

But what do you mean by maximum sum of column and row

kunal_mehta_the_great
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  • for ex. in all rows which of the row has the maximum sum, for ex. 2nd row liks this [1 2 3] has maximum sum of = some value, here its 6. – Chethu Aug 31 '16 at 14:34