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I want to unpack the tuples I create by doing the following so he the result is just one simple list. I can get the desired result in 2-3 lines but surely there is a oneliner list.comp?
x = range(10) y = [(i,j**2) for i,j in zip(x,x)] >>>y [(0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36), (7, 49), (8, 64), (9 , 81)] >>> What I want is result = [0,0,1,1,2,4,3,9.....]
Doing
y = len(x)*[0] y[::2] = x y[1::2] = [i**2 for i in x] Gives what I want but what if I need the more general case:
y = [(i, sqrt(i), i**3, some_operation_on_i, f(i), g(i)) for i in x] Eg I should be able to get a straight list like result where I only specified one operation (square) to follow each i but now with an arbitrary number of operations following each i.
740
Unpacking assigns elements of the list to multiple variables. Use the asterisk (*) in front of a variable like this *variable_name to pack the leftover elements of a list into another list.
Unpacking in Python refers to an operation that consists of assigning an iterable of values to a tuple (or list ) of variables in a single assignment statement. As a complement, the term packing can be used when we collect several values in a single variable using the iterable unpacking operator, * .
As we can see, the for loop is slower than the list comprehension (9.9 seconds vs. 8.2 seconds). List comprehensions are faster than for loops to create lists. But, this is because we are creating a list by appending new elements to it at each iteration.
Actually, list comprehension is much clearer and faster than filter+lambda, but you can use whichever you find easier. The first thing is the function call overhead: as soon as you use a Python function (whether created by def or lambda) it is likely that the filter will be slower than the list comprehension.
>>> import itertools >>> list(itertools.chain.from_iterable(y)) [0, 0, 1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81] Use a nested list comprehension:
result = [a for tup in y for a in tup] Example:
>>> x = range(10) >>> y = [(i,j**2) for i,j in zip(x,x)] >>> [a for tup in y for a in tup] [0, 0, 1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6, 36, 7, 49, 8, 64, 9, 81] This will work fine for your more general case as well, or you could do it all in one step:
y = [a for i in x for a in (i, sqrt(i), i**3, some_operation_on_i, f(i), g(i))] In case the nested list comprehensions look odd, here is how this would look as a normal for loop:
y = [] for i in x: for a in (i, sqrt(i), i**3, some_operation_on_i, f(i), g(i)): y.append(a) If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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