Question

我有一个数组数组，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，则将它们合并并将第一个元素添加在一起。因此，例如，有5个Summer Smith实例。代码应识别出同一位玩家，因此将所有Summer Smith得分相加，从而使Summer Smith总得分为20。它应该为每个玩家做到这一点。因此，一切看起来都像

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

我试图...

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

Answer 1

考虑使用dictionary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，您可以将结果作为列表列表取回：

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

Answer 2

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]]

所有defaultdict最佳：

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 3

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 4

您可以使用import java.util.*; public class HelloWorld { public static void main(String[] args) { List number = new ArrayList(); number.add(100); number.add(1000); number.add(10000); int[] changedNumber = number.stream(); int A = chagedNumber.get[2]; int B = chagedNumber.get[0]; if (A > B) { System.out.println(number.get(2) + " is bigger than " + number.get(0)); } else { System.out.println(number.get(2) + " is smaller than" + number.get(0)); } } }：

itertools.groupby

相当于循环：

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])]

并返回：

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)])

识别数组中的相似实例并将其合并

4 个答案: