列表理解中的双重迭代

Question

In Python you can have multiple iterators in a list comprehension, like

[(x,y) for x in a for y in b]

for some suitable sequences a and b. I'm aware of the nested loop semantics of Python's list comprehensions.

My question is: Can one iterator in the comprehension refer to the other? In other words: Could I have something like this:

[x for x in a for a in b]

where the current value of the outer loop is the iterator of the inner?

As an example, if I have a nested list:

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

what would the list comprehension expression be to achieve this result:

[1,2,3,4]

?? (Please only list comprehension answers, since this is what I want to find out).

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 R eturn + O uter-loop as the only R ight O rder

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 could never write double list comprehension on my first attempt. Reading into PEP202 , it turns out the reason is that it was implemented in the opposite way you would read it in English. The good news is that it is a logically sound implementation, so once you understand the structure, it's very easy to get right.

Let a, b, c, d be successively nested objects. For me, the intuitive way to extend list comprehension would mimic English:

# works
[f(b) for b in a]
# does not work
[f(c) for c in b for b in a]
[f(c) for c in g(b) for b in a]
[f(d) for d in c for c in b for b in a]

In other words, you'd be reading from the bottom up, ie

# wrong logic
(((d for d in c) for c in b) for b in a)

However this is not how Python implements nested lists. Instead, the implementation treats the first chunk as completely separate, and then chains the for s and in s in a single block from the top down (instead of bottom up), ie

# right logic
d: (for b in a, for c in b, for d in c)

Note that the deepest nested level ( for d in c ) is farthest from the final object in the list ( d ). The reason for this comes from Guido himself :

The form [... for x... for y...] nests, with the last index varying fastest, just like nested for loops.

Using Skam's text example, this becomes even more clear:

# word: for sentence in text, for word in sentence
[word for sentence in text for word in sentence]

# letter: for sentence in text, for word in sentence, for letter in word
[letter for sentence in text for word in sentence for letter in word]

# letter:
#     for sentence in text if len(sentence) > 2, 
#     for word in sentence[0], 
#     for letter in word if letter.isvowel()
[letter for sentence in text if len(sentence) > 2 for word in sentence[0] for letter in word if letter.isvowel()]

Answer 9

I feel this is easier to understand

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

result: [1, 2, 3, 4]

Answer 10

Additionally, you could use just the same variable for the member of the input list which is currently accessed and for the element inside this member. However, this might even make it more (list) incomprehensible.

input = [[1, 2], [3, 4]]
[x for x in input for x in x]

First for x in input is evaluated, leading to one member list of the input, then, Python walks through the second part for x in x during which the x-value is overwritten by the current element it is accessing, then the first x defines what we want to return.

Answer 11

This flatten_nlevel function calls recursively the nested list1 to covert to one level. Try this out

def flatten_nlevel(list1, flat_list):
    for sublist in list1:
        if isinstance(sublist, type(list)):        
            flatten_nlevel(sublist, flat_list)
        else:
            flat_list.append(sublist)

list1 = [1,[1,[2,3,[4,6]],4],5]

items = []
flatten_nlevel(list1,items)
print(items)

output:

[1, 1, 2, 3, 4, 6, 4, 5]