I have list like words = [MyOwnClass(), MyOwnClass(), None, MyOwnClass(), None]. I wanted to delete None.
I tried words.remove(None). but only worked when did it twice.
is this right way? or not, I want to know pythonic way.
class Dictionary:
words = [Word(), Word(), None, Word(), None]
def delete_None(self)
self.words.remove(None) # not works
self.words.remove(None) # now works
If you have to do multiple removes, it is better to rebuild from scratch:
words[:] = (x for x in words if x is not None)
This is a single linear operation, while every remove itself is linear as well because of the left-shifts needed for all subsequent elements.
Also note that the slice assignment words[:] = ... makes this a mutation on the original list object (just like the remove calls would be), not just a rebind of the variable whose effects may only be local. The generator (...) expression instead of a list comprehension [...] is more space efficient as it doesn't build an intermediate list, though the comprehension might be slightly faster.
For arbitrary objects, testing equality instead of identity would often be preferable:
lst[:] = (x for x in lst if x != obj_to_remove)
Or for a more functional approach:
from operator import ne # "not equal"
from functools import partial
lst[:] = filter(partial(ne, obj_to_remove), lst)
If you want to remove all the Nones, you can use filter:
words = list(filter(lambda x: x is not None, words))
I'm trying to return true if only all the previous elements are true up to the current position.
I have it set up with all function but I don't want to code it this way
def check(lightsOnOff, light):
for light in lights[:light]:
if not on:
return False
return True
count = count + 1
In general all is a useful construct to use, I can see why it looks wrong in this expression
all(list(lightsOnOff.values())[:light])
but the smelly part is actually the list(iterable)[:number] construction, which forces construction of the whole list then truncates it.
As an important aside, if lightsOnOff is a dict (not e.g. an OrderedDict) your code will be non-deterministic (see notes at bottom).
If you don't want to create a list and slice it, you can leverage itertools:
from itertools import islince
...
all(islice(lightsOnOff.values(), n))
As a frame challenge, if your dict has an order and you know the keys, you can simply rewrite it as:
all(lightsOnOff[k] for k in keys[:light])
and if your dict has keys that are ordered and e.g. integers, just use a list?
all(listOfLights[:light])
Provided you want to implement all yourself on an arbitrary list, you can do something like:
my_list = [1, 7, 2, 1, None, 2, 3]
up_to_ix = 5
def my_all(some_list, up_to_index):
for element in some_list[:up_to_index]:
if not element:
return False
return True
my_all(my_list, up_to_ix)
The function will loop through all elements in the list up to, but excluding the some_index and if it finds at least one Falsy value, will return False, otherwise True.
i am refreshing my python (2.7) and i am discovering iterators and generators.
As i understood, they are an efficient way of navigating over values without consuming too much memory.
So the following code do some kind of logical indexing on a list:
removing the values of a list L that triggers a False conditional statement represented here by the function f.
I am not satisfied with my code because I feel this code is not optimal for three reasons:
I read somewhere that it is better to use a for loop than a while loop.
However, in the usual for i in range(10), i can't modify the value of 'i' because it seems that the iteration doesn't care.
Logical indexing is pretty strong in matrix-oriented languages, and there should be a way to do the same in python (by hand granted, but maybe better than my code).
Third reason is just that i want to use generator/iterator on this example to help me understand.
Third reason is just that i want to use generator/iterator on this example to help me understand.
TL;DR : Is this code a good pythonic way to do logical indexing ?
#f string -> bool
def f(s):
return 'c' in s
L=['','a','ab','abc','abcd','abcde','abde'] #example
length=len(L)
i=0
while i < length:
if not f(L[i]): #f is a conditional statement (input string output bool)
del L[i]
length-=1 #cut and push leftwise
else:
i+=1
print 'Updated list is :', L
print length
This code has a few problems, but the main one is that you must never modify a list you're iterating over. Rather, you create a new list from the elements that match your condition. This can be done simply in a for loop:
newlist = []
for item in L:
if f(item):
newlist.append(item)
which can be shortened to a simple list comprehension:
newlist = [item for item in L if f(item)]
It looks like filter() is what you're after:
newlist = filter(lambda x: not f(x), L)
filter() filters (...) an iterable and only keeps the items for which a predicate returns True. In your case f(..) is not quite the predicate but not f(...).
Simpler:
def f(s):
return 'c' not in s
newlist = filter(f, L)
See: https://docs.python.org/2/library/functions.html#filter
Never modify a list with del, pop or other methods that mutate the length of the list while iterating over it. Read this for more information.
The "pythonic" way to filter a list is to use reassignment and either a list comprehension or the built-in filter function:
List comprehension:
>>> [item for item in L if f(item)]
['abc', 'abcd', 'abcde']
i want to use generator/iterator on this example to help me understand
The for item in L part is implicitly making use of the iterator protocol. Python lists are iterable, and iter(somelist) returns an iterator .
>>> from collections import Iterable, Iterator
>>> isinstance([], Iterable)
True
>>> isinstance([], Iterator)
False
>>> isinstance(iter([]), Iterator)
True
__iter__ is not only being called when using a traditional for-loop, but also when you use a list comprehension:
>>> class mylist(list):
... def __iter__(self):
... print('iter has been called')
... return super(mylist, self).__iter__()
...
>>> m = mylist([1,2,3])
>>> [x for x in m]
iter has been called
[1, 2, 3]
Filtering:
>>> filter(f, L)
['abc', 'abcd', 'abcde']
In Python3, use list(filter(f, L)) to get a list.
Of course, to filter a list, Python needs to iterate over it, too:
>>> filter(None, mylist())
iter has been called
[]
"The python way" to do it would be to use a generator expression:
# list comprehension
L = [l for l in L if f(l)]
# alternative generator comprehension
L = (l for l in L if f(l))
It depends on your context if a list or a generator is "better" (see e.g. this so question). Because your source data is coming from a list, there is no real benefit of using a generator here.
For simply deleting elements, especially if the original list is no longer needed, just iterate backwards:
Python 2.x:
for i in xrange(len(L) - 1, -1, -1):
if not f(L[i]):
del L[i]
Python 3.x:
for i in range(len(L) - 1, -1, -1):
if not f(L[i]):
del L[i]
By iterating from the end, the "next" index does not change after deletion and a for loop is possible. Note that you should use the xrange generator in Python 2, or the range generator in Python 3, to save memory*.
In cases where you must iterate forward, use your given solution above.
*Note that Python 2's xrange will break if there are >= 2 ** 32 - 1 elements. Python 3's range, as well as the less efficient Python 2's range do not have this limitation.
I need to retrieve a list (lets call it list1) of the double of every element smaller than 5 and put them in another list (let’s call it list2). I tried to use map as below but for some reason I get invalid syntax:
list2 = map(lambda x: x*2 if x < 5, list1)
I suspect it’s because the ternary expression needs an else condition. Is that it? And what should I do about it?
You’re right about the ternary expression part. Python doesn't allow you to use the syntax: var = <action> if <condition> without else because, in the case where <condition> == False, var becomes unknown.
You don’t really need map, however, you could use list comprehensions because not only do they solve your problems, they’re more efficient than mapping:
list2 = [x*2 for x in list1 if x < 5]
When I meet the situation I can do it in javascript, I always think if there's an foreach function it would be convenience. By foreach I mean the function which is described below:
def foreach(fn,iterable):
for x in iterable:
fn(x)
they just do it on every element and didn't yield or return something,i think it should be a built-in function and should be more faster than writing it with pure Python, but I didn't found it on the list,or it just called another name?or I just miss some points here?
Maybe I got wrong, cause calling an function in Python cost high, definitely not a good practice for the example. Rather than an out loop, the function should do the loop in side its body looks like this below which already mentioned in many python's code suggestions:
def fn(*args):
for x in args:
dosomething
but I thought foreach is still welcome base on the two facts:
In normal cases, people just don't care about the performance
Sometime the API didn't accept iterable object and you can't rewrite its source.
Every occurence of "foreach" I've seen (PHP, C#, ...) does basically the same as pythons "for" statement.
These are more or less equivalent:
// PHP:
foreach ($array as $val) {
print($val);
}
// C#
foreach (String val in array) {
console.writeline(val);
}
// Python
for val in array:
print(val)
So, yes, there is a "foreach" in python. It's called "for".
What you're describing is an "array map" function. This could be done with list comprehensions in python:
names = ['tom', 'john', 'simon']
namesCapitalized = [capitalize(n) for n in names]
Python doesn't have a foreach statement per se. It has for loops built into the language.
for element in iterable:
operate(element)
If you really wanted to, you could define your own foreach function:
def foreach(function, iterable):
for element in iterable:
function(element)
As a side note the for element in iterable syntax comes from the ABC programming language, one of Python's influences.
Other examples:
Python Foreach Loop:
array = ['a', 'b']
for value in array:
print(value)
# a
# b
Python For Loop:
array = ['a', 'b']
for index in range(len(array)):
print("index: %s | value: %s" % (index, array[index]))
# index: 0 | value: a
# index: 1 | value: b
map can be used for the situation mentioned in the question.
E.g.
map(len, ['abcd','abc', 'a']) # 4 3 1
For functions that take multiple arguments, more arguments can be given to map:
map(pow, [2, 3], [4,2]) # 16 9
It returns a list in python 2.x and an iterator in python 3
In case your function takes multiple arguments and the arguments are already in the form of tuples (or any iterable since python 2.6) you can use itertools.starmap. (which has a very similar syntax to what you were looking for). It returns an iterator.
E.g.
for num in starmap(pow, [(2,3), (3,2)]):
print(num)
gives us 8 and 9
The correct answer is "python collections do not have a foreach". In native python we need to resort to the external for _element_ in _collection_ syntax which is not what the OP is after.
Python is in general quite weak for functionals programming. There are a few libraries to mitigate a bit. I helped author one of these infixpy
pip install infixpy https://pypi.org/project/infixpy/
from infixpy import Seq
(Seq([1,2,3]).foreach(lambda x: print(x)))
1
2
3
Also see: Left to right application of operations on a list in Python 3
Here is the example of the "foreach" construction with simultaneous access to the element indexes in Python:
for idx, val in enumerate([3, 4, 5]):
print (idx, val)
Yes, although it uses the same syntax as a for loop.
for x in ['a', 'b']: print(x)
This does the foreach in python 3
test = [0,1,2,3,4,5,6,7,8,"test"]
for fetch in test:
print(fetch)
Look at this article. The iterator object nditer from numpy package, introduced in NumPy 1.6, provides many flexible ways to visit all the elements of one or more arrays in a systematic fashion.
Example:
import random
import numpy as np
ptrs = np.int32([[0, 0], [400, 0], [0, 400], [400, 400]])
for ptr in np.nditer(ptrs, op_flags=['readwrite']):
# apply random shift on 1 for each element of the matrix
ptr += random.choice([-1, 1])
print(ptrs)
d:\>python nditer.py
[[ -1 1]
[399 -1]
[ 1 399]
[399 401]]
If I understood you right, you mean that if you have a function 'func', you want to check for each item in list if func(item) returns true; if you get true for all, then do something.
You can use 'all'.
For example: I want to get all prime numbers in range 0-10 in a list:
from math import sqrt
primes = [x for x in range(10) if x > 2 and all(x % i !=0 for i in range(2, int(sqrt(x)) + 1))]
If you really want you can do this:
[fn(x) for x in iterable]
But the point of the list comprehension is to create a list - using it for the side effect alone is poor style. The for loop is also less typing
for x in iterable: fn(x)
I know this is an old thread but I had a similar question when trying to do a codewars exercise.
I came up with a solution which nests loops, I believe this solution applies to the question, it replicates a working "for each (x) doThing" statement in most scenarios:
for elements in array:
while elements in array:
array.func()
If you're just looking for a more concise syntax you can put the for loop on one line:
array = ['a', 'b']
for value in array: print(value)
Just separate additional statements with a semicolon.
array = ['a', 'b']
for value in array: print(value); print('hello')
This may not conform to your local style guide, but it could make sense to do it like this when you're playing around in the console.
In short, the functional programming way to do this is:
def do_and_return_fn(og_fn: Callable[[T], None]):
def do_and_return(item: T) -> T:
og_fn(item)
return item
return do_and_return
# where og_fn is the fn referred to by the question.
# i.e. a function that does something on each element, but returns nothing.
iterable = map(do_and_return_fn(og_fn), iterable)
All of the answers that say "for" loops are the same as "foreach" functions are neglecting the point that other similar functions that operate on iters in python such as map, filter, and others in itertools are lazily evaluated.
Suppose, I have an iterable of dictionaries coming from my database and I want to pop an item off of each dictionary element when the iterator is iterated over. I can't use map because pop returns the item popped, not the original dictionary.
The approach I gave above would allow me to achieve this if I pass lambda x: x.pop() as my og_fn,
What would be nice is if python had a built-in lazy function with an interface like I constructed:
foreach(do_fn: Callable[[T], None], iterable: Iterable)
Implemented with the function given before, it would look like:
def foreach(do_fn: Callable[[T], None], iterable: Iterable[T]) -> Iterable[T]:
return map(do_and_return_fn(do_fn), iterable)
# being called with my db code.
# Lazily removes the INSERTED_ON_SEC_FIELD on every element:
doc_iter = foreach(lambda x: x.pop(INSERTED_ON_SEC_FIELD, None), doc_iter)
No there is no from functools import foreach support in python. However, you can just implement in the same number of lines as the import takes, anyway:
foreach = lambda f, iterable: (*map(f, iterable),)
Bonus:
variadic support: foreach = lambda f, iterable, *args: (*map(f, iterable, *args),) and you can be more efficient by avoiding constructing the tuple of Nones