Question

迭代元组列表 [（＆＃39; a＆＃39;，4），（＆＃39; b＆＃39;，5），（＆＃39; c＆＃39;，1），（＆＃39; d＆＃39;， 3），（＆＃39; e＆＃39;，2），（＆＃39; f＆＃39;，6）]

尝试将匹配对作为输出

非常感谢任何指导。

这是我到目前为止所做的：

data = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
new_list = []
vowels = 'aeiou'
consonants = 'bcdfghjklmnpqrstvxw'
consonants = list(consonants)
vowels = list(vowels)
for idx, (a,b) in enumerate(data):
    if (a) in vowels or (a) in consonants and (b) % 3 == 0:
        new_list.append(idx)
print tuple(new_list)

这是我遇到的问题

Answer 1

这是一种简单的方法，您可以在以后进行扩展：

nums = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
vowels = ['a','e','i','o','u']
results = []
for k,v in enumerate(nums):
    for i,j in enumerate(nums[:]):
        if v == j:
            continue
        if v[0] in vowels and j[0] in vowels:
            if (v[1]+j[1]) % 3 == 0:
                if (i,k) not in results:
                    results.append((k,i))
        if v[0] not in vowels and j[0] not in vowels:
            if (v[1]+j[1]) % 3 == 0:
                if (k,i) not in results:
                    results.append((i,k))
print(results)

Answer 2

data = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)]
new_list = []
vowels = 'aeiou'
for idx,(a,b) in enumerate(data):
     for _idx,(_a,_b) in enumerate( data[idx+1:]):
           if bool(a in vowels)==bool(_a in vowels) and(b+_b)%3==0:
               new_list.append((idx,_idx+idx+1))

Answer 3

a = [('a', 4), ('b', 5), ('c', 1), ('d', 3), ('e', 2), ('f',6)] 
c = np.asarray([[int(e[0] in 'aeiou'), e[1]] for e in a])
e = (c[:,None] + c) % 3
f = np.asarray(np.where((e[...,0]%2==0) & (e[...,1]==0))).T.tolist()
print set([tuple(sorted(i)) for i in f if i[0]!=i[1]])

set([(1L, 2L), (3L, 5L), (0L, 4L)])

迭代元组列表

3 个答案: