Sure.
test = '01'
for m in test:
for n in test:
for o in test:
for p in test:
print m+n+o+p
[snip]
Forgive any silly mistakes I have made (I've been teaching
myself python for about 1 week) but there is a moderately
well known algorithm for this that extends to arbitrary
lengths of both the list of alternatives and the length
of the required output, and avoids deeply nested loops.
I know that it is no better for small and constant output
lengths, but for longer lengths or if the output length
can vary it should be better. There is a similar algorithm
if duplicates are not allowed (ie abcd ... wxyz).
My attempt at a python translation of the algorithm:
def m_from_n ( v, m ):
"""
Print all combinations of m things from v[0] ... v[n-1],
duplicates OK. Yields a list.
"""
x = [0] * m
while True:
yield [ v
for i in x ]
i = m - 1
while i>=0 and x==len(v)-1:
x = 0
i = i - 1
if i >= 0:
x = x + 1
else:
return
for y in m_from_n( "xyz", 2 ):
print ''.join(y)
xx
xy
xz
yx
yy
yz
zx
zy
zz
for y in m_from_n( [0,1], 3 ):
print y
[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]
for y in m_from_n( "abcdefghijklmnopqrstuvwxyz", 4 ):
print ''.join(y)
should more or less do what you want.
Charles
