How can I make a for-loop pyramid more concise in Python? [duplicate]

Based on what you want to do, you can use the itertools module to minimize the for loops (or zip).In this case itertools.product would create what you have done with the 4 loops:

>>> list(product(range(3),repeat=4))[(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 0, 2), (0, 0, 1, 0), (0, 0, 1, 1), (0, 0, 1, 2), (0, 0, 2, 0), (0, 0, 2, 1), (0, 0, 2, 2), (0, 1, 0, 0), (0, 1, 0, 1), (0, 1, 0, 2), (0, 1, 1, 0), (0, 1, 1, 1), (0, 1, 1, 2), (0, 1, 2, 0), (0, 1, 2, 1), (0, 1, 2, 2), (0, 2, 0, 0), (0, 2, 0, 1), (0, 2, 0, 2), (0, 2, 1, 0), (0, 2, 1, 1), (0, 2, 1, 2), (0, 2, 2, 0), (0, 2, 2, 1), (0, 2, 2, 2), (1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 0, 2), (1, 0, 1, 0), (1, 0, 1, 1), (1, 0, 1, 2), (1, 0, 2, 0), (1, 0, 2, 1), (1, 0, 2, 2), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 0, 2), (1, 1, 1, 0), (1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 2, 0), (1, 1, 2, 1), (1, 1, 2, 2), (1, 2, 0, 0), (1, 2, 0, 1), (1, 2, 0, 2), (1, 2, 1, 0), (1, 2, 1, 1), (1, 2, 1, 2), (1, 2, 2, 0), (1, 2, 2, 1), (1, 2, 2, 2), (2, 0, 0, 0), (2, 0, 0, 1), (2, 0, 0, 2), (2, 0, 1, 0), (2, 0, 1, 1), (2, 0, 1, 2), (2, 0, 2, 0), (2, 0, 2, 1), (2, 0, 2, 2), (2, 1, 0, 0), (2, 1, 0, 1), (2, 1, 0, 2), (2, 1, 1, 0), (2, 1, 1, 1), (2, 1, 1, 2), (2, 1, 2, 0), (2, 1, 2, 1), (2, 1, 2, 2), (2, 2, 0, 0), (2, 2, 0, 1), (2, 2, 0, 2), (2, 2, 1, 0), (2, 2, 1, 1), (2, 2, 1, 2), (2, 2, 2, 0), (2, 2, 2, 1), (2, 2, 2, 2)]

And in your code you can do :

for i,j,k,l in product(range(3),repeat=4):    #do stuff

This function is equivalent to the following code, except that the actual implementation does not build up intermediate results in memory:

def product(*args, **kwds):    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111    pools = map(tuple, args) * kwds.get('repeat', 1)    result = [[]]    for pool in pools:        result = [x+[y] for x in result for y in pool]    for prod in result:        yield tuple(prod)

Edit:As @ PeterE says in comment product() can be used even if the ranges have differing length :

product(range(3),range(4),['a','b','c'] ,some_other_iterable)

The idea to use itertools.product is a good one. Here's a more general approach that will support ranges of varying sizes.

from itertools import productdef product_of_ranges(*ns):    for t in product(*map(range, ns)):        yield tfor i, j, k in product_of_ranges(4, 2, 3):    # do stuff

It won't be more concise as it will cost you a generator function, but at least you won't be bothered by PEP8 :

def tup4(n):    for i in range(n):        for j in range(n):            for k in range(n):                for l in range(n):                    yield (i, j, k, l)for (i, j, k, l) in tup4(3):    # do your stuff

(in python 2.x you should use xrange instead of range in the generator function)

EDIT:

Above method should be fine when the depth of the pyramid is known. But you can also make a generic generator that way without any external module :

def tup(n, m):    """ Generate all different tuples of size n consisting of integers < m """    l = [ 0 for i in range(n)]    def step(i):        if i == n : raise StopIteration()        l[i] += 1        if l[i] == m:            l[i] = 0            step(i+ 1)    while True:        yield tuple(l)        step(0)for (l, k, j, i) in tup(4, 3):    # do your stuff

(I used (l, k, j, i) because in above generator, first index varies first)