如何在python中对N个列表的元素求和?

简单地说:

L = [(['a', 1], ['b', 2]), (['c', 3], ['d', 4], ['e', 5])]

for t in L:
    s = ''.join(x[0] for x in t)
    S = sum(x[1] for x in t)
    print([s, S])

并避免使用python关键字定义变量名。。。

这里有一个想法,使用嵌套列表理解 zip 和一个用于求和或求和的自定义函数:

lst = [(['a', 1], ['b', 2]), (['c', 3], ['d', 4], ['e', 5])]

def sum_or_join(l):
    try:
        return sum(l)
    except TypeError:
        return ''.join(l)
    
out = [[sum_or_join(x) for x in zip(*t)] for t in lst]

或者如果在内部元组中总是有两个项 ('string', int)

out = [[f(x) for f, x in zip((''.join, sum), zip(*t))]
       for t in lst]

输出: [['ab', 3], ['cde', 12]]

我希望这有帮助

lst = [ (['a', 1], 
         ['b', 2]), 
         
        (['c', 3], 
         ['d', 4], 
         ['e', 5]) ]

for i in list:
  print["".join([d for d in zip(*i)][0]), sum([d for d in zip(*i)][1])]

x+y ,它只适用于循环的第一次迭代,因为只有2个值需要解包。但它在第二次迭代中不起作用。

int str 在一起

• 为了避免混淆,我将重命名变量 list 到 lst

[('ab', 3), ('cde', 12)] 结果

lst = [(['a', 1], ['b', 2]), (['c', 3], ['d', 4], ['e', 5])] # don't use list 
                                                             # as variable name
# One-liner using builtin functions sum, list and Walrus operator
result = [(''.join(p[0]), sum(p[1])) for sublist in lst if (p:=list(zip(*sublist)))]

• 总和

这应该清楚明了。

A = [(['a', 1], ['b', 2]), (['c', 3], ['d', 4], ['e', 5])]

result = []
for tup in A:
    char, value = "", 0
    for x, y in tup:
        char += x
        value += y
    result.append([char, value])
print(result)

zip :

result = []
for tup in A:
    tmp = []
    for x in zip(*tup):
        if type(x[0]) == str:
            tmp.append(''.join(x))
        elif type(x[0]) == int:
            tmp.append(sum(x))
    result.append(tmp)
print(result)

样品O/P:

[['ab', 3], ['cde', 12]]

result = [[''.join([x for x,y in item]),sum([y for x,y in item])] for item in lst]

看看这个。

result = [("".join(list(zip(*tup))[0]), sum(list(zip(*tup))[1])) for tup in lst]