Double Iteration in List Comprehension

Answer #1:

I hope this helps someone else since a,b,x,y don’t have much meaning to me! Suppose you have a text full of sentences and you want an array of words.

# Without list comprehension
list_of_words = []
for sentence in text:
    for word in sentence:
       list_of_words.append(word)
return list_of_words

I like to think of list comprehension as stretching code horizontally.

Try breaking it up into:

# List Comprehension 
[word for sentence in text for word in sentence]

Example:

>>> text = (("Hi", "Steve!"), ("What's", "up?"))
>>> [word for sentence in text for word in sentence]
['Hi', 'Steve!', "What's", 'up?']

This also works for generators

>>> text = (("Hi", "Steve!"), ("What's", "up?"))
>>> gen = (word for sentence in text for word in sentence)
>>> for word in gen: print(word)
Hi
Steve!
What's
up?

Answer #2:

To answer your question with your own suggestion:

>>> [x for b in a for x in b] # Works fine

While you asked for list comprehension answers, let me also point out the excellent itertools.chain():

>>> from itertools import chain
>>> list(chain.from_iterable(a))
>>> list(chain(*a)) # If you're using python < 2.6

Answer #3:

Gee, I guess I found the anwser: I was not taking care enough about which loop is inner and which is outer. The list comprehension should be like:

[x for b in a for x in b]

to get the desired result, and yes, one current value can be the iterator for the next loop.

Answer #4:

Order of iterators may seem counter-intuitive.

Take for example: [str(x) for i in range(3) for x in foo(i)]

Let’s decompose it:

def foo(i):
    return i, i + 0.5

[str(x)
    for i in range(3)
        for x in foo(i)
]

# is same as
for i in range(3):
    for x in foo(i):
        yield str(x)

Answer #5:

ThomasH has already added a good answer, but I want to show what happens:

>>> a = [[1, 2], [3, 4]]
>>> [x for x in b for b in a]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'b' is not defined

>>> [x for b in a for x in b]
[1, 2, 3, 4]
>>> [x for x in b for b in a]
[3, 3, 4, 4]

I guess Python parses the list comprehension from left to right. This means, the first for loop that occurs will be executed first.

The second “problem” of this is that b gets “leaked” out of the list comprehension. After the first successful list comprehension b == [3, 4].

Answer #6:

This memory technic helps me a lot:

[ <RETURNED_VALUE> <OUTER_LOOP1> <INNER_LOOP2> <INNER_LOOP3> ... <OPTIONAL_IF> ]

And now you can think about Return + Outer-loop
as the only Right Order

Knowing above, the order in list comprehensive even for 3 loops seem easy:

c=[111, 222, 333]
b=[11, 22, 33]
a=[1, 2, 3]

print(
  [
    (i, j, k)                            # <RETURNED_VALUE> 
    for i in a for j in b for k in c     # in order: loop1, loop2, loop3
    if i < 2 and j < 20 and k < 200      # <OPTIONAL_IF>
  ]
)
[(1, 11, 111)]

because the above is just a:

for i in a:                         # outer loop1 GOES SECOND
  for j in b:                       # inner loop2 GOES THIRD
    for k in c:                     # inner loop3 GOES FOURTH
      if i < 2 and j < 20 and k < 200:
        print((i, j, k))            # returned value GOES FIRST

for iterating one nested list/structure, technic is the same:
for a from the question:

a = [[1,2],[3,4]]
[i2    for i1 in a      for i2 in i1]
which return [1, 2, 3, 4]

for one another nested level

a = [[[1, 2], [3, 4]], [[5, 6], [7, 8, 9]], [[10]]]
[i3    for i1 in a      for i2 in i1     for i3 in i2]
which return [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

and so on

Answer #7:

If you want to keep the multi dimensional array, one should nest the array brackets. see example below where one is added to every element.

>>> a = [[1, 2], [3, 4]]

>>> [[col +1 for col in row] for row in a]
[[2, 3], [4, 5]]

>>> [col +1 for row in a for col in row]
[2, 3, 4, 5]

Answer #8:

I feel this is easier to understand

[row[i] for row in a for i in range(len(a))]

result: [1, 2, 3, 4]