如何在python中的2D列表中比较第一个元素并操作第二个元素

Question

I have a 2D list (list of lists) in python in this format:

x=[[1,'A'],[1,'B'],[2,'A'],[3,'T'],[3,'Z'],[8,'K'],[6,'K'],[8,'N']]

I want to compare the first element that is integer and if the number is equal to another number in the list then combine their 2nd element ie alphabet. Desired Output (List of strings):

["1:'A','B'","2:'A'","3:'T','Z'","8:'K',N'","6:'K'"]

I did try to use for loop but it is not checking the similarity of more than 2 elements. Is there any other simpler method for this?

My attempt:

for i in range(len(x)):
    if x[i][0]==x[i+1][0]:
        print x[i][-1]+","+x[i+1][-1]
    else:
        print x[i][-1]

Answer 1

You can construct your strings once you group your elements as you described, and for that you can use groupby :

from itertools import groupby

a = [[1,'A'],[1,'B'],[2,'A'],[3,'T'],[3,'Z'],[8,'K'],[6,'K'],[8,'N']]

{k: [v[1] for v in g] for k, g in groupby(sorted(a), key=lambda x: x[0])}
#{1: ['A', 'B'], 2: ['A'], 3: ['T', 'Z'], 6: ['K'], 8: ['K', 'N']}

From this, as of Python 3.6, you can do the following:

r = {k: [v[1] for v in g] for k, g in groupby(sorted(a), key=lambda x: x[0])}
result = ["{0}: {','.join(repr(i) for i in {1})}".format(k, v) for k, v in r.items()]

result
#["1: 'A','B'", "2: 'A'", "3: 'T','Z'", "6: 'K'", "8: 'K','N'"]

Answer 2

Using collections.defaultdict

Ex:

from collections import defaultdict
x=[[1,'A'],[1,'B'],[2,'A'],[3,'T'],[3,'Z'],[8,'K'],[6,'K'],[8,'N']]
result = defaultdict(list)

for k, v in x:
    result[k].append(v)

result = ["{}: {}".format(k, ",".join(v)) for k,v in result.items()]
print(result)

---

Or using dict.setdefault

Ex:

result = {}
for k, v in x:
    result.setdefault(k, []).append(v)

---

Output:

['8: K,N', '1: A,B', '2: A', '3: T,Z', '6: K']