Develop Reference

How to use list slicing to append to end of list in python?

Following can be used to add a slice to append to front of list.
>>> a = [5,6]
>>> a[0:0] = [1,2,3]
>>> a
[1,2,3,5,6]
what slice to use to append to the end of list.

If you really want to use slice, you can use the length of a:
a = [5, 6]
a[len(a):] = [1, 2, 3]
a
output:
[5, 6, 1, 2, 3]
But the simplest is to directly extend a:
a = [5, 6]
a += [1, 2, 3] # or a.extend([1, 2, 3])

I think you should consider extend():
>>> a = [1, 2, 3]
>>> a.extend([4, 5, 6])
>>> a
[1, 2, 3, 4, 5, 6]
Both + and += operators are defined for list, which are semantically
similar to extend.
list + list2 creates a third list in memory, so you can return the
result of it, but it requires that the second iterable be a list.
list += list2 modifies the list in-place (it is the in-place operator,
and lists are mutable objects, as we've seen) so it does not create a
new list. It also works like extend, in that the second iterable can
be any kind of iterable.
Time Complexity
Append has constant time complexity, O(1).
Extend has time complexity, O(k).
Iterating through the multiple calls to append adds to the complexity,
making it equivalent to that of extend, and since extend's iteration
is implemented in C, it will always be faster if you intend to append
successive items from an iterable onto a list.
↳ More Information

>>> a = [1, 2, 3]
>>> a[len(a):] = [4, 5, 6]
>>> a
[1, 2, 3, 4, 5, 6]
or
>>> a = [1, 2, 3]
>>> a += [4, 5, 6]
>>> a
[1, 2, 3, 4, 5, 6]

You have got short answer from Jeevaa and Reblochon Masque but if you want to use for loop then try this:
a = [5,6]
b = [1,2,3]
for val in b[::-1]:#Reverse b and insert it to a
a.insert(0,val)
print(a)
Output
[1,2,3,5,6]

pythonic way of iterating over variable number of sequences [duplicate]

Say I have list1 = [1,2,3,4] and list2 = [5,6,7,8]. How would I compare the first element, 1, in list1 with the first element, 5, in list2? And 2 with 6, 3 with 7, and so on.
I'm trying to use a for loop for this, but I'm not sure how to do it. I understand that doing for x in list1 just checks an element x to all elements in list1, but I don't know how to do it when comparing two lists the way I described.

You can traverse both lists simultaneously using zip:
for (x, y) in zip(list1, list2): do_something
The 'zip' function gives you [(1,5), (2,6), (3,7), (4,8)], so in loop iteration N you get the Nth element of each list.

The default comparison operators compare lists in lexicographical order. So you can say things like:
>>> [1, 2, 3, 4] < [5, 6, 7, 8]
True
If instead you want to compute the elementwise comparison, you can use map and cmp (or any other operator:
>>> map(cmp, [1, 2, 3, 4], [5, 6, 7, 8])
[-1, -1, -1, -1]

If your result is going to be a new list, then you can use a list comprehension:
new_list = [ some_function(i, j) for i, j in zip(list1, list2) ]
Here's a real example of the above code:
>>> list1 = [1, 2, 3, 4]
>>> list2 = [1, 3, 4, 4]
>>> like_nums = [ i == j for i, j in zip(list1, list2) ]
>>> print like_nums
[True, False, False, True]
This will make a list of bools that show whether items of the same index in two lists are equal to each other.
Furthermore, if you use the zip function, there is a way to unzip the result when you are done operating on it. Here's how:
>>> list1 = [1, 2, 3, 4]
>>> list2 = [1, 3, 4, 4]
>>> new_list = zip(list1, list2) # zip
>>> print new_list
[(1, 1), (2, 3), (3, 4), (4, 4)]
>>> newlist1, newlist2 = zip(*new_list) # unzip
>>> print list(newlist1)
[1, 2, 3, 4]
>>> print list(newlist2)
[1, 3, 4, 5]
This could be useful if you need to modify the original lists, while also comparing the elements of the same index in some way.

function of difference between value

Is there a function in Python to get the difference between two or more values in a list? So, in those two lists:
list1 = [1, 5, 3, 7]
list2 = [4, 2, 6, 4]
I need to calculate the difference between every value in list1 and list2.
for i in list1:
for ii in list2:
print i -ii
This gives negative values, but I want the subtraction between the values of the two lists only from highest value to lowest value for not getting negative values.
For the above lists, I expect the output to be [3, 3, 3, 3].
Thanks.

Assuming you expect [3, 3, 3, 3] as the answer in your question, you can use abs and zip:
[abs(i-j) for i,j in zip(list1, list2)]

Either zip the lists, or use numpy:
>>> list1 = [1, 5, 3, 7]
>>> list2 = [4, 2, 6, 4]
>>> [a-b for a,b in zip(list1, list2)]
[-3, 3, -3, 3]
>>> import numpy as np
>>> np.array(list1) - np.array(list2)
array([-3, 3, -3, 3])
Remember to cast the array back to a list as needed.
edit:
In response to the new requirement that the absolute values are needed: you can add abs in the list comprehension:
>>> [abs(a-b) for a,b in zip(list1, list2)]
[3, 3, 3, 3]
and the numpy solution would change to:
>>> map(abs, np.array(list1) - np.array(list2))
[3, 3, 3, 3]

You may also do if else condition inside list comprehension.
>>> [i-j if i>j else j-i for i,j in zip(list1, list2)]
[3, 3, 3, 3]

You can use zip method in order combine these two lists. See the tutorials for zip method https://docs.python.org/2/library/functions.html#zip
>>> list1 = [1, 5, 3, 7]
>>> list2 = [4, 2, 6, 4]
>>> [abs(x-y) for x, y in zip(list1, list2)]
[3, 3, 3, 3]

Avinash Raj's answer is correct, or alternatively, using map():
from operator import sub
C = map(sub, A, B)

Repeat each item in a list a number of times specified in another list

I have two lists, x and y:
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
I want to use these to create a new list. The new list will have each element in x repeated the number of times specified by the corresponding element in y. Hence, the desired output is
>>> new_list
[2, 3, 3, 4, 4, 4]
The order of the elements in new_list doesn't matter to me. It's also not crucial that it be a list -- any sequence type is fine.
What is the fastest, most efficient, most Pythonic way to achieve this?

numpy's repeat function gets the job done:
>>> import numpy as np
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
>>> np.repeat(x, y)
array([2, 3, 3, 4, 4, 4])

You can use list comprehension, like this
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
>>> [item for item, count in zip(x, y) for i in range(count)]
[2, 3, 3, 4, 4, 4]
Here, we zip the x and y so that the element from x and its corresponding count from y are grouped as a single tuple. Then, we iterate count number of items to produce the same item.
If your objects in x are immutables, then you can create count copies of the same and put them together in a list, like this
>>> [i for item, count in zip(x, y) for i in [item] * count]
[2, 3, 3, 4, 4, 4]
You can do the same lazily, with itertools.repeat, like this
>>> from itertools import chain, repeat
>>> chain.from_iterable((repeat(item, count) for item, count in zip(x,y)))
<itertools.chain object at 0x7fabe40b5320>
>>> list(chain.from_iterable((repeat(item, cnt) for item, cnt in zip(x,y))))
[2, 3, 3, 4, 4, 4]
Please note that the chain returns an iterable, not a list. So, if you don't want all the elements at once, you can get the items one by one from it. This will be highly memory efficient if the count is going to be a very big number, as we don't create the entire list in the memory immediately. We generate the values on-demand.
Thanks ShadowRanger. You can actually apply repeat over x and y and get the result like this
>>> list(chain.from_iterable(map(repeat, x, y)))
[2, 3, 3, 4, 4, 4]
here, map function will apply the values from x and y to repeat one by one. So, the result of map will be
>>> list(map(repeat, x, y))
[repeat(2, 1), repeat(3, 2), repeat(4, 3)]
Now, we use chain.from_iterable to consume values from each and every iterable from the iterable returned by map.

Simple using for loop.
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
>>> final = []
>>> for index, item in enumerate(y):
final.extend([x[index]]*item)

One way to achieve this is via using .elements() function of collections.Counter() along with zip. For example:
>>> from collections import Counter
>>> x = [2, 3, 4]
>>> y = [1, 2, 3]
# `.elements()` returns an object of `itertool.chain` type, which is an iterator.
# in order to display it's content, here type-casting it to `list`
>>> list(Counter(dict(zip(x,y))).elements())
[2, 3, 3, 4, 4, 4]

Python: How can I obtain something in a list of list?

Currently I have the following list:
>>> a = list()
>>> b = [1, 2, 3]
>>> c = [4,5,6]
>>> d = [7, 8]
>>> a.append(b)
>>> a.append(c)
>>> a.append(d)
>>> del b, c, d
>>> print a
[[1, 2, 3], [4, 5, 6], [7, 8]]
I want to obtain the first item on b, c and d (value 1, 4 and 7). How could I do that in one line of code?
I want to obtain the remaining item on b, c and d (value 2, 3, 5, 6 and 8). How could I do that in one line of code?

>>> zip(*a)[0]
(1, 4)
The zip(*a) is like a transpose function.
>>> zip(*a)
[(1, 4), (2, 5), (3, 6)]
The [0] indexes the first element.
>>> zip(*a)[0]
(1, 4)
The second problem is trickier - zip won't do the trick. Instead you need a list comprehension.
>>> [i for l in a for i in l[1:]]
[2, 3, 5, 6, 8]
Or if you don't need a flattened list, simply:
>>> [l[1:] for l in a]
[[2, 3], [5, 6], [8]]

This should do the trick:
[lst[0] for lst in a]
It iterates over a and for every element (which is a list) it collects the first value.

There are plenty of ways to do this. This would be one:
>>> print [x[0] for x in a]
[1,4]
Or putting print inside the for loop:
>>> for x in a: print x[0],
1 4
The most basic way if you only want the first elements of the first two sublists:
>>> print a[0][0], a[1][0]
1 4

I'd go with niklas B. answer zip(b,c)[0], but, to give you more flexibility, if you don't know the number of sublists in the list, you can also try:
[el[0] for el in a]
and gives you all the first items in the sublists of a.