识别数组中的类似实例并合并它们

Question

我有一个数组数组，如下。。。

[['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，则将它们合并并将第一个元素添加到一起。例如，Summer Smith有5个实例。代码应该识别出是同一名球员，因此将Summer Smith的所有得分相加，使Summer Smith总得分为20。它应该为每个球员做到这一点。例如，

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

我试着…

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

Answer 1

考虑使用字典来跟踪每个玩家的分数总和：

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

然后你可以以列表的形式返回结果：

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

Answer 2

我会使用Counter

>>> 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})

Answer 3

更好地使用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]]

最佳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]]

Answer 4

您可以使用itertools.groupby进行以下操作：

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]]