關閉名單的名單，分組特定元素上附加其他元素

Question

我怎麼能轉換的列表，如：

l=[ ['A', 'C21'], ['A','D43'],['B','D34'],['C','D45'],['C',D56'] 

到：

[ ['A','C21 D43'], ['B','D34'],['C','D45 D56'] ] 

凡分組根據元素進行每個子列表 的＃0和元素＃1是在每個組內的字符串連接？

Answer 1

試試這個：

l=[ ['A', 'C21'], ['A','D43'],['B','D34'],['C','D45'],['C','D56']] 
x = {} 

for a in l: 
    if a[0] not in x.keys(): 
     x[a[0]] = [a[1]] 
    else: 
     x[a[0]].append(a[1]) 

print x 

array_result = [] 
for keys, vals in x.iteritems(): 
    array_result.append([keys, ' '.join(vals)]) 

print array_result 

Answer 2

如果密鑰是連續的，那麼你可以使用itertools.groupby，如：

from itertools import groupby 

data =[ ['A', 'C21'], ['A','D43'],['B','D34'],['C','D45'],['C','D56'] ] 
new_data = [[k, ' '.join(el[1] for el in g)] for k, g in groupby(data, lambda L: L[0])] 
# [['A', 'C21 D43'], ['B', 'D34'], ['C', 'D45 D56']] 

如果不是和訂單其實並不重要，那麼：

from collections import defaultdict 

dd = defaultdict(list) 
for key, val in data: 
    dd[key].append(val) 

new_data = [[k, ' '.join(v)] for k,v in dd.items()] 
# [['B', 'D34'], ['C', 'D45 D56'], ['A', 'C21 D43']] 

或者 - 使用dict.setdefault，例如：

d = {} 
for key, val in data: 
    d.setdefault(key, []).append(val) 
new_data = [[k, ' '.join(v)] for k,v in d.items()] 

或者，如果密鑰是不連續的，但輸出應該保持輸入的順序，然後用collections.OrderedDict，如：

from collections import OrderedDict 

d = OrderedDict() 
for key, val in data: 
    d.setdefault(key, []).append(val) 

new_data = [[k, ' '.join(v)] for k,v in d.items()] 
# [['A', 'C21 D43'], ['B', 'D34'], ['C', 'D45 D56']]