What is the analogue of Haskell's zipWith function in Python?
zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
map()
map(operator.add, [1, 2, 3], [3, 2, 1])
Although a LC with zip() is usually used.
[x + y for (x, y) in zip([1, 2, 3], [3, 2, 1])]
You can create yours, if you wish, but in Python we mostly do
list_c = [ f(a,b) for (a,b) in zip(list_a,list_b) ]
as Python is not inherently functional. It just happens to support a few convenience idioms.
You can use map:
>>> x = [1,2,3,4]
>>> y = [4,3,2,1]
>>> map(lambda a, b: a**b, x, y)
[1, 8, 9, 4]
A lazy zipWith with itertools:
import itertools
def zip_with(f, *coll):
return itertools.starmap(f, itertools.izip(*coll))
This version generalizes the behaviour of zipWith with any number of iterables.
Generally as others have mentioned map and zip can help you replicate the functionality of zipWith as in Haskel.
Generally you can either apply a defined binary operator or some binary function on two list.An example to replace an Haskel zipWith with Python's map/zip
Input: zipWith (+) [1,2,3] [3,2,1]
Output: [4,4,4]
>>> map(operator.add,[1,2,3],[4,3,2])
[5, 5, 5]
>>> [operator.add(x,y) for x,y in zip([1,2,3],[4,3,2])]
[5, 5, 5]
>>>
There are other variation of zipWith aka zipWith3, zipWith4 .... zipWith7. To replicate these functionalists you may want to use izip and imap instead of zip and map.
>>> [x for x in itertools.imap(lambda x,y,z:x**2+y**2-z**2,[1,2,3,4],[5,6,7,8],[9,10,11,12])]
>>> [x**2+y**2-z**2 for x,y,z in itertools.izip([1,2,3,4],[5,6,7,8],[9,10,11,12])]
[-55, -60, -63, -64]
As you can see, you can operate of any number of list you desire and you can still use the same procedure.
I know this is an old question, but ...
It's already been said that the typical python way would be something like
results = [f(a, b) for a, b in zip(list1, list2)]
and so seeing a line like that in your code, most pythonistas will understand just fine.
There's also already been a (I think) purely lazy example shown:
import itertools
def zipWith(f, *args):
return itertools.starmap(f, itertools.izip(*args))
but I believe that starmap returns an iterator, so you won't be able to index, or go through multiple times what that function will return.
If you're not particularly concerned with laziness and/or need to index or loop through your new list multiple times, this is probably as general purpose as you could get:
def zipWith(func, *lists):
return [func(*args) for args in zip(*lists)]
Not that you couldn't do it with the lazy version, but you could also call that function like so if you've already built up your list of lists.
results = zipWith(func, *lists)
or just like normal like:
results = zipWith(func, list1, list2)
Somehow, that function call just looks simpler and easier to grok than the list comprehension version.
Looking at that, this looks strangely reminiscent of another helper function I often write:
def transpose(matrix):
return zip(*matrix)
which could then be written like:
def transpose(matrix):
return zipWith(lambda *x: x, *matrix)
Not really a better version, but I always find it interesting how when writing generic functions in a functional style, I often find myself going, "Oh. That's just a more general form of a function I've already written before."
Related
So, basically I've got a few functions that return tuples. Essentially of the form:
def function():
return (thing, other_thing)
I want to be able to add several of these functions together in a straightforward way, like this:
def use_results(*args):
"""
Each arg is a function like the one above
"""
results = [test() for test in args]
things = magic_function(results)
other_things = magic_function(results)
Basically I have the data structure:
[([item_1, item_1], [item_2, item_2]), ([item_3, item_3], [item_4, item_4])]
and I want to turn it into:
[[item_1, item_1, item_3, item_3], [item_2, item_2, item_4, item_4]]
It seems like there's probably a nice pythonic way of doing this with a combination of zip and *, but it's not quite coming to me.
Oh, I feel kind of silly. I found an answer quickly after posting the question. I'm going to still keep this up in case there's a better solution though:
>>> import operator
>>> results = [([1,1], [2,2]), ([3,3], [4,4])]
>>> map(operator.add, *results)
[[1, 1, 3, 3], [2, 2, 4, 4]]
Without importing any module, just built-in methods:
>>> results = [([1,1], [2,2]), ([3,3], [4,4])]
>>> [x+y for x,y in zip(*results)]
[[1, 1, 3, 3], [2, 2, 4, 4]]
Or even this way as well:
>>> map(lambda s,t:s+t, *results)
In Python what is equivalent to Ruby's Array.each method? Does Python have a nice and short closure/lambda syntax for it?
[1,2,3].each do |x|
puts x
end
Does Python have a nice and short closure/lambda syntax for it?
Yes, but you don't want it in this case.
The closest equivalent to that Ruby code is:
new_values = map(print, [1, 2, 3])
That looks pretty nice when you already have a function lying around, like print. When you just have some arbitrary expression and you want to use it in map, you need to create a function out of it with a def or a lambda, like this:
new_values = map(lambda x: print(x), [1, 2, 3])
That's the ugliness you apparently want to avoid. And Python has a nice way to avoid it: comprehensions:
new_values = [print(x) for x in values]
However, in this case, you're just trying to execute some statement for each value, not accumulate the new values for each value. So, while this will work (you'll get back a list of None values), it's definitely not idiomatic.
In this case, the right thing to do is to write it explicitly—no closures, no functions, no comprehensions, just a loop:
for x in values:
print x
The most idiomatic:
for x in [1,2,3]:
print x
You can use numpy for vectorized arithmetic over an array:
>>> import numpy as np
>>> a = np.array([1, 2, 3])
>>> a * 3
array([3, 6, 9])
You can easily define a lambda that can be used over each element of an array:
>>> array_lambda=np.vectorize(lambda x: x * x)
>>> array_lambda([1, 2, 3])
array([1, 4, 9])
But as others have said, if you want to just print each, use a loop.
There are also libraries that wrap objects to expose all the usual functional programming stuff.
PyDash http://pydash.readthedocs.org/en/latest/
underscorepy (Google github underscore.py)
E.g. pydash allows you to do things like this:
>>> from pydash import py_
>>> from __future__ import print_function
>>> x = py_([1,2,3,4]).map(lambda x: x*2).each(print).value()
2
4
6
8
>>> x
[2, 4, 6, 8]
(Just always remember to "trigger" execution and/or to un-wrap the wrapped values with .value() at the end!)
without need of an assignment:
list(print(_) for _ in [1, 2, 3])
or just
[print(_) for _ in [1, 2, 3]]
So my friend presented a problem for me to solve, and I'm currently writing a solution in functional-style Python. The problem itself isn't my question; I'm looking for a possible idiom that I can't find at the moment.
What I need is a fold, but instead of using the same function for every one of it's applications, it would do a map-like exhaustion of another list containing functions. For example, given this code:
nums = [1, 2, 3]
funcs = [add, sub]
special_foldl(nums, funcs)
the function (special_foldl) would fold the number list down with ((1 + 2) - 3). Is there a function/idiom that elegantly does this, or should I just roll my own?
There is no such function in the Python standard library. You'll have to roll you own, perhaps something like this:
import operator
import functools
nums = [1, 2, 3]
funcs = iter([operator.add, operator.sub])
def special_foldl(nums, funcs):
return functools.reduce(lambda x,y: next(funcs)(x,y), nums)
print(special_foldl(nums, funcs))
# 0
Normally, list comprehensions are used to derive a new list from an existing list. Eg:
>>> a = [1, 2, 3, 4, 5]
>>> [i for i in a if i > 2]
[3, 4, 5]
Should we use them to perform other procedures? Eg:
>>> a = [1, 2, 3, 4, 5]
>>> b = []
>>> [b.append(i) for i in a]
[None, None, None, None, None]
>>> print b
[1, 2, 3, 4, 5]
or should I avoid the above and use the following instead?:
for i in a:
b.append(i)
You should indeed avoid using list comprehensions (along with dictionary comprehensions, set comprehensions and generator expressions) for side effects. Apart from the fact that they'd accumulate a bogus list and thus waste memory, it's also confusing. I expect a list comprehension to generate a (meaningful) value, and many would agree. Loops, on the other hand, are clearly a sequence of statements. They are expected to kick off side effects and generate no result value - no surprise.
From python documentation:
List comprehensions provide a concise way to create lists. Common
applications are to make new lists
Perhaps you want to learn more about reduce(), filter() and map() functions.
In the example you give it would make the most sense to do:
b = [i for i in a]
if for some reason you wanted to create b. In general, there is some common sense that must be employed. If using a comprehension makes your code unreadable, don't use it. Otherwise go for it.
Only use list comprehensions if you plan to use the created list. Otherwise you create it just for the GC to throw it again without ever being used.
So instead of [b.append(i) for i in a] you should use a proper for loop:
for i in a:
b.append(i)
Another solution would be through a generator expression:
b += (i for i in a)
However, if you want to append the whole list, you can simply do
b += a
And if you just need to apply a function to the elements before adding them to the list, you can always use map:
b += map(somefunc, a)
b = []
a = [1, 2, 3, 4, 5]
b.extend (a)
Is there an equivalent of cons in Python? (any version above 2.5)
If so, is it built in? Or do I need easy_install do get a module?
WARNING AHEAD: The material below may not be practical!
Actually, cons needs not to be primitive in Lisp, you can build it with λ.
See Use of lambda for cons/car/cdr definition in SICP for details. In Python, it is translated to:
def cons(x, y):
return lambda pair: pair(x, y)
def car(pair):
return pair(lambda p, q: p)
def cdr(pair):
return pair(lambda p, q: q)
Now, car(cons("a", "b")) should give you 'a'.
How is that? Prefix Scheme :)
Obviously, you can start building list using cdr recursion. You can define nil to be the empty pair in Python.
def nil(): return ()
Note that you must bind variable using = in Python. Am I right? Since it may mutate the variable, I'd rather define constant function.
Of course, this is not Pythonic but Lispy, not so practical yet elegant.
Exercise: Implement the List Library http://srfi.schemers.org/srfi-1/srfi-1.html of Scheme in Python. Just kidding :)
In Python, it's more typical to use the array-based list class than Lisp-style linked lists. But it's not too hard to convert between them:
def cons(seq):
result = None
for item in reversed(seq):
result = (item, result)
return result
def iter_cons(seq):
while seq is not None:
car, cdr = seq
yield car
seq = cdr
>>> cons([1, 2, 3, 4, 5, 6])
(1, (2, (3, (4, (5, (6, None))))))
>>> iter_cons(_)
<generator object uncons at 0x00000000024D7090>
>>> list(_)
[1, 2, 3, 4, 5, 6]
Note that Python's lists are implemented as vectors, not as linked lists. You could do lst.insert(0, val), but that operation is O(n).
If you want a data structure that behaves more like a linked list, try using a Deque.
In Python 3, you can use the splat operator * to do this concisely by writing [x, *xs]. For example:
>>> x = 1
>>> xs = [1, 2, 3]
>>> [x, *xs]
[1, 1, 2, 3]
If you prefer to define it as a function, that is easy too:
def cons(x, xs):
return [x, *xs]
You can quite trivially define a class that behaves much like cons:
class Cons(object):
def __init__(self, car, cdr):
self.car = car
self.cdr = cdr
However this will be a very 'heavyweight' way to build basic data structures, which Python is not optimised for, so I would expect the results to be much more CPU/memory intensive than doing something similar in Lisp.
No. cons is an implementation detail of Lisp-like languages; it doesn't exist in any meaningful sense in Python.