将列表转换为字典并添加重复的键值

Question

我试图找到这样的字典值

{'deepak': 0, 'nayak': 0}

我试过这种方式

d={}

d['deepak'] = 0

d['nayak'] = 0 

f = [ '1, deepak, 15',

'2, nayak, 10',

 '3, deepak, 10',

 '4, nayak, 13']

for lines in f:

print(lines)

##appropriate logic to excute#

    print(d)

预期的 o/p 应该是

{'deepak': 25, 'nayak': 23}

Answer 1

你几乎在那里d = {}f = [ '1, deepak, 15','2, nayak, 10', '3, deepak, 10', '4, nayak, 13']for line in f:    # by splitting on the comma and trailing space    # you can unpack those three entries into a throw-away var,    # k, and v and you only need to convert v to int    _, k, v = line.split(', ')    d[k] = d.get(k, 0) + int(v)d{'deepak': 25, 'nayak': 23}

Answer 2

d={}d['deepak'] = 0d['nayak'] = 0 f = [ '1, deepak, 15','2, nayak, 10', '3, deepak, 10', '4, nayak, 13']for lines in f:    print(lines)    ##appropriate logic to excute#    line = lines.split(', ')    d[line[1]] += int(line[-1])print(d)

Answer 3

您可以Counter从collections模块中使用，以减少使用常规 dict 时所需的一些样板文件from collections import Counterf = [    '1, deepak, 15',    '2, nayak, 10',    '3, deepak, 10',    '4, nayak, 13']result = Counter()for x in f:    _, name, count = x.split(', ')    result[name] += int(count)print(result)