Make one list of two constraints applying the list
I have two lists A, B
A = [2,3,1,4,5,2,4]
B = [4,2,3,6,2,5,1]
I want to combine these A and B as:
C = [2,4,2,3,1,3,4,6,2,5,2,5,1,4]
regulations:
- A [i] <= B [i], take A [i], then B [i]
- A [i]> B [i], take B [i], then A [i]
I can do this using a loop like below:
C = []
for a,b in zip(A,B):
if(a<=b):
C.append(a)
C.append(b)
else:
C.append(b)
C.append(a)
It really works. How can I do it like this:
C = [ [a,b if (a<=b)],[ b,a else] for a,b in zip(A,B)] # This is totally wrong
But how can I do this using if-else
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The way you do it is good because it is very readable ... but if the one-liner is what you commit after me:
>>> A = [2,3,1,4,5,2,4]
>>> B = [4,2,3,6,2,5,1]
>>> [i for sublist in [[a, b] if a < b else [b, a] for a, b in zip(A, B)] for i in sublist]
[2, 4, 2, 3, 1, 3, 4, 6, 2, 5, 2, 5, 1, 4]
A few notes:
-
When you add a conditional expression to the comp list, use
if - else
immediately after the first variable in the comp list.['a' if i in (2, 4, 16) else 'b' for i in [1, 2, 3, 16, 24]]
-
The best way to build (mentally) nested lists is to think about how you would write it in a normal loop.
C = [[a, b] if a < b else [b, a] for a, b in zip(A, B)]
for sublist in C:
for i in sublist:
yield i
Then you just flatten the nested loops and bring them yield i
to the front, discarding yield
.
for sublist in C for i in sublist yield i
|-> yield i for sublist in C for i in sublist
|-> i for sublist in C for i in sublist
Now you can just replace C with the comp list above and get the one-liner I posted.
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Use itertools.chain.from_iterable
,
import itertools
A = [2,3,1,4,5,2,4]
B = [4,2,3,6,2,5,1]
list(itertools.chain.from_iterable(i if i[0]<=i[1] else (i[1], i[0]) for i in zip(A, B)))
or collect tuple
. You get by running two listszip(A, B)
list(itertools.chain.from_iterable(sorted(i) for i in zip(A, B)))
See. Time difference, using List Comprehension
, map
,sorted
In [70]: %timeit list(itertools.chain.from_iterable(i if i[0]<=i[1] else (i[1], i[0]) for i in zip(A, B)))
100000 loops, best of 3: 3.49 µs per loop
In [71]: %timeit list(itertools.chain.from_iterable(sorted(i) for i in zip(A, B)))
100000 loops, best of 3: 5.81 µs per loop
In [72]: %timeit [i for sublist in [[a, b] if a < b else [b, a] for a, b in zip(A, B)] for i in sublist]
100000 loops, best of 3: 3.28 µs per loop
In [73]: %timeit list(itertools.chain.from_iterable(map(lambda x:x[1]>x[0] and (x[0],x[1]) or (x[1],x[0]),zip(A,B))))
100000 loops, best of 3: 4.26 µs per loop
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