我有一个输出的元组列表
[('city 1', '#' ), ('city 1', '#' ), ('city 2', '#' ), ('city 2', '#'), etc...]
我需要做的是将每个城市的所有 # 值相加并输出一个列表:
[('city 1', 'sumof#' ), ('city 2', 'sumof#' ), ('city 3', 'sumof#' ), ('city 4', 'sumof#'), etc...]
请您参考如下方法:
不确定你想要什么。可能是这个？:
>>> d = [('city 1', 2 ), ('city 1', 1 ), ('city 2', 2 ), ('city 2', 4)]
# Have an empty dictionary to calculate and store result
>>> r = {}
# Iterate through the list
# As the list contains tuples with 2 elements, we can just
# make use of unpacking
>>> for city,count in d:
# dict.setdefault will try to get the city count, but
# if not exists (first time) it will return the provided
# default value; 0 here. Just add up.
r[city] = r.setdefault(city,0) + count
# items() contains the key-value pair. May be this is the result?
>>> r.items()
[('city 2', 6), ('city 1', 3)]
