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Cartesian product of two or more lists: Difference between revisions

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Cartesian product of two or more lists (view source)

Revision as of 16:31, 30 January 2019

978 bytes removed ,  5 years ago
→‎{{header|Python}}: Remove non-idiomatic, convoluted, " general applicative function" version.
(→‎{{header|Python}}: Replaced original with a function wrapper around itertools.product)
(→‎{{header|Python}}: Remove non-idiomatic, convoluted, " general applicative function" version.)
Line 2,105:
[]
</pre>
 
 
and we could define our own '''cartesianProduct''', in terms of a general applicative function:
 
<lang python>from functools import (reduce)
 
 
# cartesianProduct :: [a] -> [b] -> [(a, b)]
def cartesianProduct(xs):
return lambda ys: ap(
map(lambda x: lambda y: (x, y), xs)
)(ys)
 
 
# GENERIC FUNCTIONS --------------------------------
 
# apList (<*>) :: [(a -> b)] -> [a] -> [b]
def ap(fs):
return lambda xs: reduce(
lambda a, f: a + reduce(
lambda a, x: a + [f(x)], xs, []
), fs, []
)
 
 
# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
def g(x, y):
return f(x)(y)
return g
 
 
# TEST ------------------------------------------------
for product in map(lambda x: uncurry(cartesianProduct)(*x), ([
([1, 2], [3, 4]),
([3, 4], [1, 2]),
([1, 2], []),
([], [1, 2])
])):
print (product)</lang>
{{Out}}
<pre>[(1, 3), (1, 4), (2, 3), (2, 4)]
[(3, 1), (3, 2), (4, 1), (4, 2)]
[]
[]</pre>
 
=={{header|R}}==
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