Python 识别数组中的相似实例并合并它们

我有一个数组，像这样

[['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4]]

我希望能够扫描每个数组，如果数组中已经存在元素0，则合并它们并将第一个元素添加到一起。例如，Summer Smith有5个例子。代码应该识别它是同一个玩家，因此将所有Summer Smith分数相加，使Summer Smith总分数为20。它应该对每个玩家都这样做。所以这一切看起来像是，比如

[['Harry',20], ['Jake', 16]]....

我试着

for array in arrays:
        if array[0] in [not sure what to do now]

考虑使用a来跟踪每个玩家的得分总和：

scores = [['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4]]

grouped_scores = {}
for name, score in scores:
    if name not in grouped_scores: grouped_scores[name] = score
    else: grouped_scores[name] += score

然后，您可以将结果作为列表列表返回：

merged_scores = [list(t) for t in grouped_scores.items()]
print(merged_scores)
# [['Summer Smith', 20], ['Scary Terry', 20], ['Abradolf Lincler', 32]]

---

您可以为此使用

itertools.groupby

：

from itertools import groupby
l = [['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4]]

res = [[name, sum(s[1] for s in score)] for name, score in groupby(sorted(l, key=lambda x: x[0]), key=lambda x: x[0])]

这相当于循环：

res = []
for name, score in groupby(sorted(l, key=lambda x: x[0]), key=lambda x: x[0]):
    res.append([name, sum(s[1] for s in score)])

并返回：

>>> res
[['Abradolf Lincler', 32], ['Scary Terry', 20], ['Summer Smith', 20]]

---

最好使用

setdefault

：

d={}
for x,y in arrays:
   d.setdefault(x,[]).append(y)
print(list(map(list,{k:sum(v) for k,v in d.items()}.items())))

输出：

[['Abradolf Lincler', 32], ['Scary Terry', 20], ['Summer Smith', 20]]

[['Abradolf Lincler', 32], ['Scary Terry', 20], ['Summer Smith', 20]]

最好的

defaultdict

：

from collections import defaultdict
d=defaultdict(int)
for x,y in arrays:
   d[x]+=y
print(list(map(list,dict(d).items())))   

输出：

[['Abradolf Lincler', 32], ['Scary Terry', 20], ['Summer Smith', 20]]

[['Abradolf Lincler', 32], ['Scary Terry', 20], ['Summer Smith', 20]]

---

我会使用

计数器

>>> from collections import Counter
>>> arrays = [['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Scary Terry', 4], ['Scary Terry', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Summer Smith', 4], ['Abradolf Lincler', 4], ['Summer Smith', 4], ['Summer Smith', 4]]
>>> result = Counter()
>>> for k, v in arrays:
...     result[k] += v
...
>>> result
Counter({'Abradolf Lincler': 32, 'Scary Terry': 20, 'Summer Smith': 20})

---

你有列表，而不是数组。你有的是列表。数组通常是

bytearray

或

array.array

（stdlib）或

numpy.array

（第三方）。列表就是列表。更好：

d[x]+=y

，但你是对的：

defaultdict

是最好的。@jpp哦，是的，谢谢：-），编辑完毕，也谢谢你说

defaultdict

是最好的：d