Need to assign each tuple in a list to a value in a dict

Question

I want to assign a value to each tuple in a list. Let's say my list looks like this:

[(68, 125), (113, 69), (65, 86), (108, 149), (152, 53), (78, 90), (54, 160), (20, 137), (107, 90), (48, 12)]

I need to assign each of these tuples into a dict where it should look like:

1 [(48, 12)]
2 [(68, 125), (54, 160), (20, 137)]
3 [(113, 69), (152, 53)]
4 [(108, 149)]

This is what I have so far:

new_dict = {"1":[], "2":[], "3":[], "4":[]}
for (x,y) in numbers:
    if (x,y) == (48, 12):
        new_dict += {"1:"[(x,y)]}
    elif (x,y) == (68, 125) or (x,y) == (54, 160) or (x,y) == (20, 137):
        new_dict += {"2:"[(x,y)]}
    elif (x,y) == (113, 69) or (x,y) == (152, 53):
        new_dict += {"3:"[(x,y)]}
    elif (x,y) == (108, 149):
        new_dict += {"4:"[(x,y)]}
    return new_dict

Help will be much appreciated. Thank you.

It is part of some homework which I have to do. I just need to know how to do this part here. I can upload the rest of this if you want but it is not relevant. — R. Mercy, Jun 02 '16 at 11:49

score 1 · Accepted Answer · 2016-06-02T11:59:06.710

1

There are some problems in your code.

iterate over the tuples, you don't have to split them
compare using in, it's shorter than using or
use append to add to an existing list

Try this:

new_dict = {"1":[], "2":[], "3":[], "4":[]}
for t in numbers:
    if t == (48, 12):
        new_dict["1"].append(t)
    elif t in [(68, 125), (54, 160), (20, 137)]:
        new_dict["2"].append(t)
    elif t in [(113, 69), (152, 53)]:
        new_dict["3"].append(t)
    elif t == (108, 149):
        new_dict["4"].append(t)
print(new_dict)

Output:

{'1': [(48, 12)], '2': [(68, 125), (54, 160), (20, 137)], '4': [(108, 149)], '3': [(113, 69), (152, 53)]}

edited Jun 02 '16 at 11:59

answered Jun 02 '16 at 11:46

Thank you for this. However this give me the wrong output: 4 [] 1 [] 2 [(68, 125)] 3 [] – R. Mercy Jun 02 '16 at 11:53
Sure, because it is a non generic question. But I like your answer. – Jun 02 '16 at 12:06
4 [] 1 [] 2 [(68, 125)] 3 [] – R. Mercy Jun 02 '16 at 12:55
@R.Mercy The code from my answer produces the output I've included. Don't use any `return`, just try it as it is. – Jun 02 '16 at 12:57
I tried the new edited version but it randomises the order of the values. – R. Mercy Jun 02 '16 at 13:27
A dict has no order. – Jun 02 '16 at 13:28
Is there any way to sort it – R. Mercy Jun 02 '16 at 14:08
see https://stackoverflow.com/questions/9001509/how-can-i-sort-a-dictionary-by-key – Jun 02 '16 at 14:12

score 1 · Answer 2 · answered Jun 02 '16 at 12:03

1

If you know the group of each element (as you mentioned), you can proceed with:

from itertools import groupby

L  = [(68, 125), (113, 69), (65, 86), (108, 149), (152, 53), (78, 90), (54, 160), (20, 137), (107, 90), (48, 12)]
ix = [1, 2, 2, 2, 3, 3, 4]

{i:map(lambda x: x[0], g) for i, g in groupby(zip(L,ix), key=lambda x: x[1])}

Out[216]:
{1: [(68, 125)],
 2: [(113, 69), (65, 86), (108, 149)],
 3: [(152, 53), (78, 90)],
 4: [(54, 160)]}

answered Jun 02 '16 at 12:03

Colonel Beauvel

30,423
11
47
87

again, who downvote without reading question/answers? seems common in python forum ... – Colonel Beauvel Jun 02 '16 at 12:06
2

User is a noobie, throwing map, zip, and groupby at her isn't going to help. In fact, many people (myself included) absolutely hate map and zip since they are complex, so they obfuscate code, and the Python interpreter can iterate just as fast via explicit for loops. So, I yank map and zip whenever I see them in a codebase I'm working on if at all possible - the noobs on my team will scratch heads, panic, or misinterpret what they're doing. – Kevin J. Rice Jun 02 '16 at 12:09
1

or do not employ noobs :) It's also by twicking these functions that people progress a little bit in programming ... – Colonel Beauvel Jun 02 '16 at 12:10
Both are valid arguments. How senior does a Python developer have to be to understand such basic functional programming? – Jun 02 '16 at 12:58

Need to assign each tuple in a list to a value in a dict

2 Answers2