使元组列表中的元素在 python 中唯一的有效方法是什么？

Question

Let's say I have a list of tuples

l = [('A', 12345), ('A', 2435), ('A', 2342), ('B', 2968), ('B', 9483), ('C', 563)]

What is the most efficient way to make the items in my list unique, like this:

l = [('A.1', 12345), ('A.2', 2435), ('A.3', 2342), ('B.1', 2968), ('B.2', 9483), ('C.1', 563)]

Answer 1

One approach could be to group with itertools.groupby() and then "expand" the groups:

from itertools import groupby
from operator import itemgetter

l = [('A', 12345), ('A', 2435), ('A', 2342), ('B', 2968), ('B', 9483), ('C', 563)]


print([
    (f'{k}.{index}', v) 
    for k, g in groupby(l, itemgetter(0)) 
    for index, (_, v) in enumerate(g, start=1)
])

Prints:

[('A.1', 12345), ('A.2', 2435), ('A.3', 2342), ('B.1', 2968), ('B.2', 9483), ('C.1', 563)]

Note that for the grouping to work, the input l needed to be ordered by the grouping key which it seems is the case for this example input.

Answer 2

By request, i've posted a pandas approach to this question:

import pandas as pd
df =  pd.DataFrame(l)

# Create a count per group and add them to the string:
df[0] = df[0] + "." + list(map(str,list(df.groupby(0).cumcount()+1)))

# Transpose the columns to rows so we can aggregate by 2 and create a tuple:
df.T.groupby(np.arange(len(df.T))//2).agg(tuple).to_numpy().tolist()[0]

Output

[('A.1', 12345),
 ('A.2', 2435),
 ('A.3', 2342),
 ('B.1', 2968),
 ('B.2', 9483),
 ('C.1', 563)]

Answer 3

You could also group with a collections.defaultdict() , then flatten the result with itertools.chain.from_iterable() . This works whether the result is sorted or not.

from collections import defaultdict

from itertools import chain

l = [("A", 12345), ("A", 2435), ("A", 2342), ("B", 2968), ("B", 9483), ("C", 563)]

# First group by first item in tuple
groups = defaultdict(list)
for k, v in l:
    groups[k].append(v)
# defaultdict(<class 'list'>, {'A': [12345, 2435, 2342], 'B': [2968, 9483], 'C': [563]})

# Now flatten grouped items into a flat list
result = list(
    chain.from_iterable(
        (("%s.%d" % (k, i), e) for i, e in enumerate(v, start=1))
        for k, v in groups.items()
    )
)

print(result)

Output:

[('A.1', 12345), ('A.2', 2435), ('A.3', 2342), ('B.1', 2968), ('B.2', 9483), ('C.1', 563)]