Python: using map() without lambdas on a multidimensional array/list

Question

I have some code in python, that bitwise or-equals b to all the values in an a multidimensional list called a

for i in xrange(len(a)):
    for j in xrange(len(a[i])):
        a[i][j] |= b

My question is, is there any way to write this code using only (map(), filter(), reduce()) without having to use lambdas or any other function definitions like in the example below

map(lambda x: map(lambda y: y | b, x), a)

What is the use case for using the higher-order functions map, filter, and reduce without a function parameter? — Adam Mihalcin, Mar 05 '12 at 04:23

score 5 · Accepted Answer · answered Mar 05 '12 at 04:27

5

I see absolutely no reason why one should ever avoid lambdas or list comprehensions, but here goes:

import operator,functools
a = map(functools.partial(map, functools.partial(operator.or_, b)), a)

answered Mar 05 '12 at 04:27

phihag

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Note that `functools.partial` is similar to `lambda`, so IMHO the correct answer is *No you can't*, as @senderle already wrote: http://stackoverflow.com/a/9561602/1763602 . About `functools.partial` vs `lambda`: http://stackoverflow.com/a/3252425/1763602 – Marco Sulla Feb 24 '16 at 10:39

score 4 · Answer 2 · answered Mar 05 '12 at 04:21

4

map, filter, and reduce all take functions (or at least callables -- i.e. anything with a __call__ method) as arguments. So basically, no. You have to define a function, or a class.

answered Mar 05 '12 at 04:21

senderle

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...or, to state the obvious, use a built-in function that is defined elsewhere. – senderle Mar 05 '12 at 04:59

score 1 · Answer 3 · answered Mar 05 '12 at 04:31

1

Unfortunately Python has no terse currying syntax, so you can't do something like map(b |, x).

I would just use list comprehensions:

[y | b for x in a for y in x]

answered Mar 05 '12 at 04:31

Daniel Lubarov

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Python: using map() without lambdas on a multidimensional array/list

3 Answers3