如何根据Python中第一个相同的元素对3个相同大小的排序列表求和？

Question

我有一个python dictionary包含3个lists键'时间','电源'和'使用'.所有列表都具有相同数量的元素,并且所有列表都已排序.我想要做的是总结列表'power'和'usage'的所有元素,它们的索引对应于列表'time'中的相同值,以便每个时间单元只有一个功率和使用样本.

例如,转换这个字典:

{'time': [1, 2, 2, 3, 4, 4, 5],
 'power': [2, 2, 3, 6, 3, 3, 2],
 'usage': [0, 1, 1, 2, 1, 4, 7]}

到这一个:

{'time': [1, 2, 3, 4, 5],
 'power': [2, 5, 6, 6, 2],
 'usage': [0, 2, 2, 5, 7]}

我已经编写了这段有用的代码,但我不太喜欢它:

d = {'time':[1,2,2,3,4,4,5], 'power':[0,1,1,2,1,4,7], 'usage':[2,2,3,6,3,3,2]}
prev = -1
new_d = {'time':[], 'power': [], 'usage':[]}
indexes =  range( len(d['time']) )

for i in indexes:
  if d['time'][i]!=prev:
    new_d['time'].append(d['time'][i])
    new_d['power'].append(d['power'][i])
    new_d['usage'].append(d['usage'][i])
  else:
    last_power = len( new_d['power'] ) - 1
    last_usage = len( new_d['usage'] ) - 1
    new_d['power'][last_power]+=d['power'][i]
    new_d['usage'][last_usage]+=d['usage'][i]
  prev=d['time'][i]

print d
print new_d

是否有一种蟒蛇方式可以更简单,更全面地完成这项工作？

Answer 1

>>> from itertools import groupby
>>> from operator import itemgetter
>>> d = {'usage': [0, 1, 1, 2, 1, 4, 7], 'power': [2, 2, 3, 6, 3, 3, 2], 'time': [1, 2, 2, 3, 4, 4, 5]}
>>> groups = groupby(zip(d['time'], d['power'], d['usage']), key=itemgetter(0))
>>> lists = zip(*[[k] + map(sum, zip(*g)[1:]) for k, g in groups])
>>> dict(zip(('time', 'power', 'usage'), lists))
{'usage': (0, 2, 2, 5, 7), 'power': (2, 5, 6, 6, 2), 'time': (1, 2, 3, 4, 5)}

对于可变数量的键，我添加了keys变量以避免重写它们：

>>> from itertools import groupby
>>> from operator import itemgetter
>>> keys = ('time', 'power', 'usage')
>>> groups = groupby(zip(*[d[k] for k in keys]), key=itemgetter(0))
>>> lists = zip(*[[k] + map(sum, zip(*g)[1:]) for k, g in groups])
>>> dict(zip(keys, lists))
{'usage': (0, 2, 2, 5, 7), 'power': (2, 5, 6, 6, 2), 'time': (1, 2, 3, 4, 5)}