将元素保持在列表中的位数相加达到一定值

Question

From a list find out those values whose addition makes 9

aList=[81,26,27,19,108]
output, 81(because: 8+1),27(because: 2+7) and 108 (because:1+0+8) 

I tried 2 approaches:

1st approach: I could not find a way to get the value, just get the sum

s=[81,18]  
sum=0
for l in s:
    while l:
            l,dig=divmod(l,10)
            sum=sum+dig

print(sum)

2nd approach: Nasty one indeed. Take individual values from list, convert to string to separate them and again convert to int.

s=[81]  #9

sum=0

for item in s: #81
    item=str(item) # 81 to string so I can get 8 and 1
    for i in item:
        while i:
            i =int(i)
            i,dig=divmod(i,10)
            sum=sum+dig

print(sum,item)

Problem: In both cases it only works when I have single value in the list. When I have more than 1 value aList=[81,18] it gives me sum of those 2.

I would appreciate some hints/ideas on this one. Thanks in advance.

Answer 1

You could use the following list comprehension:

l = [81,26,27,19,108]
[i for i in l if sum(int(d) for d in str(i)) == 9]
# [81, 27, 108]

Which is equivalent to the following for loop:

res = []
for i in aList:
    temp = []
    for d in str(i):
        temp.append(int(d))
    if sum(temp) == 9:
        res.append(i)