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How can i sum up all values with the same index in a dictionary which each key has a nested list as a value?

I have a dictionary, each key of dictionary has a list of list (nested list) as its value. What I want is imagine we have:
x = {1: [[1, 2], [3, 5]], 2: [[2, 1], [2, 6]], 3: [[1, 5], [5, 4]]}
My question is how can I access each element of the dictionary and concatenate those with same index: for example first list from all keys:
[1,2] from first keye +
[2,1] from second and
[1,5] from third one
How can I do this?

You can access your nested list easily when you're iterating through your dictionary and append it to a new list and the you apply the sum function.
Code:
x={1: [[1,2],[3,5]] , 2:[[2,1],[2,6]], 3:[[1,5],[5,4]]}
ans=[]
for key in x:
ans += x[key][0]
print(sum(ans))
Output:
12

Assuming you want a list of the first elements, you can do:
>>> x={1: [[1,2],[3,5]] , 2:[[2,1],[2,6]], 3:[[1,5],[5,4]]}
>>> y = [a[0] for a in x.values()]
>>> y
[[1, 2], [2, 1], [1, 5]]
If you want the second element, you can use a[1], etc.

The output you expect is not entirely clear (do you want to sum? concatenate?), but what seems clear is that you want to handle the values as matrices.
You can use numpy for that:
summing the values
import numpy as np
sum(map(np.array, x.values())).tolist()
output:
[[4, 8], [10, 15]] # [[1+2+1, 2+1+5], [3+2+5, 5+6+4]]
concatenating the matrices (horizontally)
import numpy as np
np.hstack(list(map(np.array, x.values()))).tolist()
output:
[[1, 2, 2, 1, 1, 5], [3, 5, 2, 6, 5, 4]]

As explained in How to iterate through two lists in parallel?, zip does exactly that: iterates over a few iterables at the same time and generates tuples of matching-index items from all iterables.
In your case, the iterables are the values of the dict. So just unpack the values to zip:
x = {1: [[1, 2], [3, 5]], 2: [[2, 1], [2, 6]], 3: [[1, 5], [5, 4]]}
for y in zip(*x.values()):
print(y)
Gives:
([1, 2], [2, 1], [1, 5])
([3, 5], [2, 6], [5, 4])

Can't index multiple elements in list of lists in Python (using the : operator)

I found it strange that indexing using range(:) operator for list of lists is not supported.
Sometimes this result in strange values :
a = [[1, 2], [3, 4], [5, 6], [7, 8]]
>>> a
[[1, 2], [3, 4], [5, 6], [7, 8]]
>>> a[0][1]
2
>>> a[1][1]
4
>>> a[2][1]
6
However,
>>> a[0:3][1]
[3, 4]
I was expecting [2,4,6]. What am I missing here ?
I tried this on Numpy arrays as well.enter code here
>>> a
[[1, 2], [3, 4], [5, 6], [7, 8]]
>>> a[0][1]
2
>>> a[1][1]
4
>>> a[2][1]
6
>>> a[0:3][1]
[3, 4]
I know I can use list comprehension, but my question is whether ":" is supported for list of lists?

numpy arrays do support slicing, but you're not considering the shape of the array. In numpy, this array has shape:
import numpy as np
a = np.array([[1, 2], [3, 4], [5, 6], [7, 8]])
print(a.shape)
>>>(4, 2)
meaning it's 4x2. If you slice [0:3] you're returning the first three elements of the 1st dimension. i.e.:
print(a[0:3])
>>>[[1 2]
[3 4]
[5 6]]
this output has shape:
print(a[0:3].shape)
>>>(3, 2)
if you do:
print(a[0:3][1])
>>>[3 4]
You are again calling the first element of the first dimension of the array that has shape (3, 2).
Instead you want to call:
print(a[0:3][:,1])
>>>[2 4 6]
Which gives you all of the row elements (i.e. all three elements of the first dimension) at column index 1 (where 0 and 1 represent the indexes for the two dimensions of the second dimension).
even cleaner (recommended):
print(a[0:3, 1])
>>>[2 4 6]

Using : is totally supported. Explained below...
So we start with:
a = [[1, 2], [3, 4], [5, 6], [7, 8]]
You asked about:
a[0:3][1]
We want the items from list a, from positions zero to three [0:3]. Those items returned are
[1, 2] --- position 0
[3, 4] --- position 1
[5, 6] --- position 2
[7, 8] --- position 3
Then we request from that list the item in position 1, which returns:
[3, 4]
If you want to access items inside that smaller list you need to add another index, like this:
a[0:3][1][1]
would return:
4
Diagram of basic string splitting:
Your first bracket (represented in blue) is saying "give me elements in list a between positions 0 and 3, which in this case, is ALL of them.
Your second bracket (represented in red) is saying "of the results of my first bracket, give me the element that is in position 1", which is the entire sub-list [3,4]
In this specific case
a[0:3][1]
could have simply been written as
a[1]

let us assume a list of list
list=[[1,2],[3,4],[5,6],[7,8]]
then,
list[0:3]
will return a list with elements(which are also list) from index 0 to 2
[[1, 2], [3, 4], [5, 6]]
so according list[0:3][1] will return the second element([3,4]) whose index is "1" .
a[0:3][1] will not return[2,4,6] , it returns the list of list with 3 element and chooses the second element.

When you call a[0:3] the result of that is a list with the first three elements of a. You then call a[0:3][1] which returns the 2nd element of that list which is the list [3,4].

Ordinary Python lists do not support this kind of slicing.
You can get [2, 4, 6] with Numpy:
>>> import numpy as np
>>> a = np.array([[1, 2], [3, 4], [5, 6], [7, 8]])
>>> a[0:3, 1]
array([2, 4, 6])

a = [[1, 2], [3, 4], [5, 6], [7, 8]]
a[0:3]
the output of this is a list:
>>> [[1, 2], [3, 4], [5, 6]]
Therefore:
a[0:3][1]
Accesses the element at index 1, which is [3, 4]
To get the desired output from your list, use list comprehension:
[x[1] for x in a[0:3]]
>>> [2, 4, 6]

Python Pandas from dictionary

I have a dictionary
x={'XYZ': [4, 5, 6], 'ABC': [1, 2, 3]}
I want a pd.DataFrame like this:
'SomeColumnName'
'XYZ' [4,5,6]
'ABC' [1,2,3]
Whatever I do, it splits the list of x.values() in 3 separate columns. I could do a '~'.join before creating the Dataframe. Just wondering if there was an easier way

Why don't you just input the data as:
x={'XYZ': [[4, 5, 6]], 'ABC': [[1, 2, 3]]}
Then you get:
In [7]: pd.DataFrame(x).transpose()
Out[7]:
0
ABC [1, 2, 3]
XYZ [4, 5, 6]
You can recode your dictionary using:
for key in x.keys():
x[key] = [x[key]]

Ok, this is how I did it
z = pd.DataFrame.from_records(list(x.items()),columns=['A','SomeColumnName'],index='A')
Problem was - I wasnt using list() for data

Adding an element to a nested python list

I am learning Python and am having trouble with lists. I want to create a list containing a nested list e.g. a = [1, [2]]. I then want to 'append' 4 to the nested list making a[1,[2,4]] and then insert 3 between [2,4] making a[1,[2, 3, 4]].
a.append(4) does [1,[2],4], I cannot find a way to achieve what I want.

Try:
>>> a = [1, [2]]
>>> a[-1].append(4)
>>> a
[1, [2, 4]]
>>> a[-1].insert(1, 3)
>>> a
[1, [2, 3, 4]]

How to retrieve list(s) that contains specific query items

I am trying to group list of items relevant to a query item. Below is an example of the problem and my attempt at it:
>>> _list=[[1,2,3],[2,3,4]]
>>> querylist=[1,2,4]
>>> relvant=[]
>>> for x in querylist:
for y in _list:
if x in y:
relvant.append(y)
My output:
>>> relvant
[[1, 2, 3], [1, 2, 3], [2, 3, 4], [2, 3, 4]]
Desired output:
[[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]],[[2, 3, 4]]]
The issue is after each loop of a query item, I expected the relevant lists to be grouped but that isn't the case with my attempt.
Thanks for your suggestions.

I think it's clearer to use a list comprehension:
>>> _list = [[1,2,3],[2,3,4]]
>>> querylist = [1,2,4]
>>> [[l for l in _list if x in l] for x in querylist]
[[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]], [[2, 3, 4]]]
The inner expression [l for l in _list if x in l] describes the list of all sublists that contain x. The outer expression's job is to get that list for all values of x in the query list.

By making minimal changes in the code provided you can create new dummy list to store values and at end of each inner loop iteration you just append it to the main list.
_list=[[1,2,3],[2,3,4]]
querylist=[1,2,4]
relvant=[]
for x in querylist:
dummy = []
for y in _list:
if x in y:
dummy.append(y)
relvant.append(dummy)
print relvant
>>> [[[1, 2, 3]], [[1, 2, 3], [2, 3, 4]],[[2, 3, 4]]]