列表重复操作

Question

我对python非常陌生，我现在真的迷路了。如果有人能帮助我，我会非常感激。我有一个列表:

list1 = [((a, b), 2), ((a, b), 5), ((c, d), 1)), ((e, f), 2), ((e, f), 4)]

我要找的输出是:

output = [((a, b), 7), ((c, d), 1), ((e, f), 6)]

我试着把它放进字典

new_dict = {i: j for i, j in list1}

但是它给了我一个错误

也许还有其他方法?

Answer 1

在代码注释中查找解释

list1 = [(('a', 'b'), 2), (('a', 'b'), 5), (('c', 'd'), 1), (('e', 'f'), 2), (('e', 'f'), 4)]
# let's create an empty dictionary
output = {}
# ('a', 'b') is a tuple and tuple is hashable so we can use it as dictionary key
# iterate over the list1
for i in list1:
# for each item check if i[0] exist in output
if i[0] in output:
# if yes just add i[1]
output[i[0]] += i[1]
else:
# create new key
output[i[0]] = i[1]

# finally print the dictionary
final_output = list(output.items())
print(final_output)

[(('a', 'b'), 7), (('c', 'd'), 1), (('e', 'f'), 6)]

Answer 2

您可以这样使用{}.get:

list1 = [(('a', 'b'), 2), (('a', 'b'), 5), (('c', 'd'), 1), (('e', 'f'), 2), (('e', 'f'), 4)]
di={}
for t in list1:
di[t[0]]=di.get(t[0],0)+t[1]
>>> di
{('a', 'b'): 7, ('c', 'd'): 1, ('e', 'f'): 6}

也可以使用Counter:

from collections import Counter 
c=Counter({t[0]:t[1] for t in list1})
>>> c
Counter({('a', 'b'): 5, ('e', 'f'): 4, ('c', 'd'): 1})

然后将其中任何一个转换为元组列表(如您所拥有的)，您使用list和{}.items():

>>> list(c.items())
[(('a', 'b'), 5), (('c', 'd'), 1), (('e', 'f'), 4)]

Answer 3

list1 = [(('a', 'b'), 2), (('a', 'b'), 5), (('c', 'd'), 1), (('e', 'f'), 2), (('e', 'f'), 4)]
sorted_dict = {}
for ele in list1:
if ele[0] in sorted_dict:
sorted_dict[ele[0]] += ele[1]
else:
sorted_dict[ele[0]] = ele[1]
print(sorted_dict)