CODEWITHSUNDEEP
pythonlistdictionary
Sayantan Ghosh
Sayantan Ghosh

Reputation: 338

How to convert a two element list into dictionary (one element will be key and other will be value), when duplicates are present?

So, I have a list of lists in the following manner.

ab_list = [['AB','35'],['AB','78'],['CD','98'],['CD','41'],['CD','67'].....]

Now, I want to convert this list into a dictionary such that it has the following form:

ab_dict = {'AB':'35|78','CD':'98|41|67'.........}

So, if the second element of the list is repeated it should be joined by pipeline and that will become the value. I am not getting the logic of how to do this. The code I am trying is:

for val in ab_list:
for value in final_list:
if val[0] == value[0]:
value[1] = value[1] + "|" + val[1]

But this is not working properly. How to do this?

Upvotes: 1

Views: 92

Answers (4)

Diptangsu Goswami
Diptangsu Goswami

Reputation: 5965

ab_list = [['AB', '35'], ['AB', '78'], ['CD', '98'], ['CD', '41'], ['CD', '67']]
ab_dict = {}

for k, v in ab_list:
    if k in ab_dict:
        ab_dict[k] += '|' + v
    else:
        ab_dict[k] = v

print(ab_dict)

Output:

{'AB': '35|78', 'CD': '98|41|67'}

Upvotes: 3

inmate37
inmate37

Reputation: 1238

Try this one:

# Stage-1
#
lst = [['AB', '35'], ['AB', '78'], ['CD', '98']]
tmp = []

for i in lst:
    tmp.append(i[0])

tmp = list(set(tmp))

# Stage-2
#
dct = {}

for i in tmp:
    dct.update({i: []})

for i in lst:
    dct[i[0]].append(i[1])

# Stage-3
#
final = {}

for i, j in dct.items():
    string = ''

    if len(j) == 1:
        string = j[0]
    else:
        for k in j:
            string += k + ' | '

        string = string[:-3]

    final.update({i: string})

Result:

{'AB': '35 | 78', 'CD': '98'}

Upvotes: 1

javidcf
javidcf

Reputation: 59701

EDIT: The solution below only works if all pairs with the same first element are consecutive, which is the case in the example but may not be in general (otherwise you would need to sort the list first to make it work, which would not be a very good solution).

This is one way to do it with groupby:

from itertools import groupby

ab_list = [['AB', '35'], ['AB', '78'], ['CD', '98'], ['CD', '41'], ['CD', '67']]
ab_dict = {k: '|'.join(it[1] for it in v) for k, v in groupby(ab_list, lambda it: it[0])}
print(ab_dict)
# {'AB': '35|78', 'CD': '98|41|67'}

Upvotes: 1

Rakesh
Rakesh

Reputation: 82765

This is one approach using setdefault and a simple iteration.

Ex:

ab_list = [['AB','35'],['AB','78'],['CD','98'],['CD','41'],['CD','67']]
result = {}
for k, v in ab_list:
    result.setdefault(k, []).append(v)

ab_dict = {k: "|".join(v) for k, v in result.items()}
print(ab_dict)

Output:

{'AB': '35|78', 'CD': '98|41|67'}

Upvotes: 6

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