Pythonic way to combine for-loop and if-statement

Question

I know how to use both for loops and if statements on separate lines, such as:

>>> a = [2,3,4,5,6,7,8,9,0]
... xyz = [0,12,4,6,242,7,9]
... for x in xyz:
...     if x in a:
...         print(x)
0,4,6,7,9

And I know I can use a list comprehension to combine these when the statements are simple, such as:

print([x for x in xyz if x in a])

But what I can't find is a good example anywhere (to copy and learn from) demonstrating a complex set of commands (not just "print x") that occur following a combination of a for loop and some if statements. Something that I would expect looks like:

for x in xyz if x not in a:
    print(x...)

Is this just not the way python is supposed to work?

Answer 1

You can use generator expressions like this:

gen = (x for x in xyz if x not in a)

for x in gen:
    print(x)

Answer 2

As per The Zen of Python (if you are wondering whether your code is "Pythonic", that's the place to go):

• Beautiful is better than ugly.
• Explicit is better than implicit.
• Simple is better than complex.
• Flat is better than nested.
• Readability counts.

The Pythonic way of getting the sorted intersection of two sets is:

>>> sorted(set(a).intersection(xyz))
[0, 4, 6, 7, 9]

Or those elements that are xyz but not in a:

>>> sorted(set(xyz).difference(a))
[12, 242]

But for a more complicated loop you may want to flatten it by iterating over a well-named generator expression and/or calling out to a well-named function. Trying to fit everything on one line is rarely "Pythonic".

---

Update following additional comments on your question and the accepted answer

I'm not sure what you are trying to do with enumerate, but if a is a dictionary, you probably want to use the keys, like this:

>>> a = {
...     2: 'Turtle Doves',
...     3: 'French Hens',
...     4: 'Colly Birds',
...     5: 'Gold Rings',
...     6: 'Geese-a-Laying',
...     7: 'Swans-a-Swimming',
...     8: 'Maids-a-Milking',
...     9: 'Ladies Dancing',
...     0: 'Camel Books',
... }
>>>
>>> xyz = [0, 12, 4, 6, 242, 7, 9]
>>>
>>> known_things = sorted(set(a.iterkeys()).intersection(xyz))
>>> unknown_things = sorted(set(xyz).difference(a.iterkeys()))
>>>
>>> for thing in known_things:
...     print 'I know about', a[thing]
...
I know about Camel Books
I know about Colly Birds
I know about Geese-a-Laying
I know about Swans-a-Swimming
I know about Ladies Dancing
>>> print '...but...'
...but...
>>>
>>> for thing in unknown_things:
...     print "I don't know what happened on the {0}th day of Christmas".format(thing)
...
I don't know what happened on the 12th day of Christmas
I don't know what happened on the 242th day of Christmas

Answer 3

The following is a simplification/one liner from the accepted answer:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]

for x in (x for x in xyz if x not in a):
    print(x)

12
242

Notice that the generator was kept inline. This was tested on python2.7 and python3.6 (notice the parens in the print ;) )

It is honestly cumbersome even so: the x is mentioned four times.

Answer 4

I personally think this is the prettiest version:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in filter(lambda w: w in a, xyz):
  print x

Edit

if you are very keen on avoiding to use lambda you can use partial function application and use the operator module (that provides functions of most operators).

https://docs.python.org/2/library/operator.html#module-operator

from operator import contains
from functools import partial
print(list(filter(partial(contains, a), xyz)))

Answer 5

I would probably use:

for x in xyz: 
    if x not in a:
        print(x...)

Answer 6

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]  
set(a) & set(xyz)  
set([0, 9, 4, 6, 7])

Answer 7

I liked Alex's answer, because a filter is exactly an if applied to a list, so if you want to explore a subset of a list given a condition, this seems to be the most natural way

mylist = [1,2,3,4,5]
another_list = [2,3,4]

wanted = lambda x:x in another_list

for x in filter(wanted, mylist):
    print(x)

this method is useful for the separation of concerns, if the condition function changes, the only code to fiddle with is the function itself

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

for x in filter(wanted, mylist):
    print(x)

The generator method seems better when you don't want members of the list, but a modification of said members, which seems more fit to a generator

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.5 for x in mylist if wanted(x))

for x in generator:
    print(x)

Also, filters work with generators, although in this case it isn't efficient

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.9 for x in mylist)

for x in filter(wanted, generator):
    print(x)

But of course, it would still be nice to write like this:

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

# for x in filter(wanted, mylist):
for x in mylist if wanted(x):
    print(x)

Answer 8

You can use generators too, if generator expressions become too involved or complex:

def gen():
    for x in xyz:
        if x in a:
            yield x

for x in gen():
    print x

Answer 9

Use intersection or intersection_update

• intersection :

  a = [2,3,4,5,6,7,8,9,0]
  xyz = [0,12,4,6,242,7,9]
  ans = sorted(set(a).intersection(set(xyz)))
  

• intersection_update:

  a = [2,3,4,5,6,7,8,9,0]
  xyz = [0,12,4,6,242,7,9]
  b = set(a)
  b.intersection_update(xyz)
  

  then b is your answer

Answer 10

based on the article here: https://towardsdatascience.com/a-comprehensive-hands-on-guide-to-transfer-learning-with-real-world-applications-in-deep-learning-212bf3b2f27a I used the following code for the same reason and it worked just fine:

an_array = [x for x in xyz if x not in a]

This line is a part of the program! this means that XYZ is an array which is to be defined and assigned previously, and also the variable a

Using generator expressions (which is recommended in the selected answer) makes some difficulties because the result is not an array

Answer 11

A simple way to find unique common elements of lists a and b:

a = [1,2,3]
b = [3,6,2]
for both in set(a) & set(b):
    print(both)

Answer 12

Well, it's possible to do it in just one line.

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
print('\n'.join([str(x) for x in xyz if x not in a]))
-----------------------------------------------------------------------
12
242

This gives you: