This question already has answers here:
Element-wise addition of 2 lists?
(17 answers)
Closed 7 years ago.
I'm sure there is a good way to accomplish what i want without looping over lists and creating new objects. Here is what I have
a = [1, 2, 3, 4]
b = [2, 3, 4, 5]
What I am looking to do is take each set of lists and sum each placeholder so that the output is
[3, 5, 7, 9]
Thoughts?
you should use zip function and list comprehension
a = [1, 2, 3, 4]
b = [2, 3, 4, 5]
[sum(t) for t in zip(a,b)]
Use numpy
import numpy as np
a = np.array([1, 2, 3, 4])
b = np.array([2, 3, 4, 5])
a+b
>>> array([3, 5, 7, 9])
Related
This question already has answers here:
Removing duplicates in lists
(56 answers)
One-liner to remove duplicates, keep ordering of list [duplicate]
(6 answers)
Closed 3 years ago.
I have a big list of strings appearing many times and I want a list of the same strings to appear only once.
An example with numbers would be:
a = [1, 2, 2, 3, 4, 4]
and I want to get
b = [1, 2, 3, 4]
What I tried is something like:
a = [1, 2, 2, 3, 4, 4]
[x for x in a if a.count(x) == 1]
[1, 3]
but this omits the duplicate numbers and takes only those appearing once.
You can try this:
import collections
a = [1, 2, 2, 2, 3, 3, 4, 4, 5, 6, 7, 7, 8]
print([item for item, count in collections.Counter(a).items()])
This question already has answers here:
What do ellipsis [...] mean in a list?
(5 answers)
Closed 5 years ago.
x = [4, 5, 6]
li = [1, 2, 3, 7]
li.insert(3,x)
x+=li
print(x)
The output is:
[4, 5, 6, 1, 2, 3, [...], 7]
I'm new to python/coding and I don't know what these ellipses are but when I do other code it starts getting weird. Wasn't sure what to ask since I have no clue what's going on. Thank you!
you're inserting a list inside your list, probably not what you want.
Then when doing this
x+=li
the representation of the list then shows an ellipsis because you're referencing the list from itself (x is referenced in li already)
To insert several items at once in a list in-place you could use slice assignment:
>>> x = [4, 5, 6]
>>> li = [1, 2, 3, 7]
>>> li[3:3] = x
>>> li
[1, 2, 3, 4, 5, 6, 7]
This question already has answers here:
Python list rotation [duplicate]
(4 answers)
Closed 8 years ago.
Suppose I have a list u = [1, 2, 3, 4, 5], and u[1:] returns [2, 3, 4, 5].
I wonder what indexing returns [2, 3, 4, 5, 1], going from the second position to the last and then the first?
You can make a general function that does this at any point in your list, just by adding two slices. This was an intentional design as to why slicing is half-open (includes left index, but excludes right index)
def rotate(l, i):
return l[i:] + l[:i]
>>> u = [1, 2, 3, 4, 5]
>>> rotate(u, 1)
[2, 3, 4, 5, 1]
>>> rotate(u, 2)
[3, 4, 5, 1, 2]
This question already has answers here:
Compact way to assign values by slicing list in Python
(5 answers)
Closed 2 years ago.
Is there a fast way to get the 1st, 3rd and 5th element from an array in Python like a[0,2,4]? Thanks.
Using operator.itemgetter:
>>> lst = [1,2,3,4,5,6,7]
>>> import operator
>>> get135 = operator.itemgetter(0, 2, 4)
>>> get135(lst)
(1, 3, 5)
You could just do this, a simple method with no imports necessary:
>>> a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]
>>> [a[i] for i in (0, 2, 4)]
[1, 3, 5]
Slicing is the simplest way to do this. You'll want to slice it with [0:5:2].
>>> range(100)[0:5:2]
[0, 2, 4]
This is the equivalent of saying "Starting from element 0, up to (but not including) element 5, give me every 2nd element."
You can use slicing to get this.
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
d = a[0:5:2]
print d
[1, 3, 5]
If you want to generalize to every other entry you would use
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
b = a[::2]
print b
[1, 3, 5, 7, 9]
You can use ,
Slicing operation on list.
>>> a=[i for i in range(10)]
>>> a[::2]
Ouput:
[0, 2, 4, 6, 8]
perhaps:
[list[0], list[2], list[4]]
This question already has answers here:
How do I make a flat list out of a list of lists?
(34 answers)
Closed 9 years ago.
I have one list like:
n = [[1, 2, 3], [4, 5, 6, 7, 8, 9]]
I want to create a function that takes a single list (see above) and concatenates all the sublists that are part of it into a single list.
n = [[1, 2, 3], [4, 5, 6, 7, 8, 9]]
nn = [ x for y in n for x in y]
>>> lst = [[1, 2, 3], [4, 5, 6, 7, 8, 9]]
>>> from itertools import chain
>>> list(chain.from_iterable(lst))
[1, 2, 3, 4, 5, 6, 7, 8, 9]
For completeness, here is a very short way to write this
>>> sum(n, [])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
But although it is tempting, you shouldn't because it has quadratic performance. ie a new list is created as each term is added, and all the previous items will be copied over and over
It's ok to use list.extend though
reduce(lambda x,y: x.extend(y) or x, n, [])
You can also concatenate by doing simply:
print n[0]+n[1]
In general this would be:
def concatenate(list):
x=[]
for i in list:
x+=i
return x
But this is not particularly efficent, just quite straightforward for a beginner.