Eli Harold Dan Green

一种可能的方法是:

import pandas as pd
    
flat_list = ['hello,5', 'mellow,4', 'mellow,2', 'yellow,2', 'yellow,7', 'hello,7', 'mellow,7', 'hello,7']
new_list = [v.split(',') for v in flat_list]
    
df = pd.DataFrame(new_list)
df[1] = df[1].astype(int)
df2 = df.groupby(0).sum()
print(df2)

输出:

    0        1
    hello   19
    mellow  13
    yellow   9

您可以非常简单地执行此操作,而无需导入其他模块,如:

t = ['hello,5', 'mellow,4', 'mellow,2', 'yellow,2', 'yellow,7', 'hello,7', 'mellow,7', 'hello,7']

d = {}
for s in t: #for each string
    w, n = s.split(',') #get the string and the number
    d[w] = d[w] + int(n) if w in d.keys() else int(n) #add the number (sum)

l = list(d.items()) #make the result a list of tuples
print(output)

输出:

[('hello', 19), ('mellow', 13), ('yellow', 9)]

下面是一个变体,用于使用 defaultdict :

from collections import defaultdict

t = ['hello,5', 'mellow,4', 'mellow,2', 'yellow,2',
     'yellow,7', 'hello,7', 'mellow,7', 'hello,7']

count = defaultdict(int)

for name_number in t:
    name, number = name_number.split(",")
    count[name] += int(number)

你也可以使用 Counter :

from collections import Counter

count = Counter()

for name_number in t:
    name, number = name_number.split(",")
    count[name] += int(number)

在这两种情况下,您都可以将输出转换为 list 属于 tuple s使用:

list(count.items())
# -> [('hello', 19), ('mellow', 13), ('yellow', 9)]

我运行了你的代码,得到了正确的结果(尽管不是你想要的格式)。