带有嵌套列表的字典理解，防止覆盖

Question

我正在尝试使用使用嵌套列表的字典理解。只要钥匙是唯一的，一切正常。但是，如果存在多个密钥，我想将值附加到该密钥而不是覆盖该值。这可能使用理解吗？

seq1 = [[1, [1,2,3,4]], [2, [5,6,7]]] 
seq2 = [[1, [1,2,3,4]], [1, [5,6,7]]] 

print {key: [val] for key, val in seq1} # Or dict(seq1) 
>>> {1: [[1, 2, 3, 4]], 2: [[5, 6, 7]]} 

print {key: [val] for key, val in seq2} 
>>> {1: [[5, 6, 7]]} # First value is overwritten 

# Desired output: 
def index_reads(reads): 
    result = {} 
    for i in reads: 
     d = dict([i]) 
     for key, val in d.iteritems(): 
      if key in result: 
       result[key].append(val) 
      else: 
       result[key] = [val] 
    return result 

print index_reads(seq1) 
>>> {1: [[1, 2, 3, 4]], 2: [[5, 6, 7]]} 

print index_reads(seq2) 
>>> {1: [[1, 2, 3, 4], [5, 6, 7]]} 

对不起，我找不到这个问题的重复。

Answer 1

你可以使用groupby从itertools。

import itertools 
import operator 

seq1 = [[1, [1,2,3,4]], [2, [5,6,7]]] 
seq2 = [[1, [1,2,3,4]], [1, [5,6,7]]] 

def index_reads(seq): 
    return {k: [i[1] for i in g] for k, g in itertools.groupby(seq, operator.itemgetter(0))} 

print index_reads(seq1) 
print index_reads(seq2) 

输出

{1: [[1, 2, 3, 4]], 2: [[5, 6, 7]]} 
{1: [[1, 2, 3, 4], [5, 6, 7]]} 

Answer 2

你不需要列表理解。作为一个更Python的方式，你可以使用dict.setdefault()方法：

>>> d={key: [val] for key, val in seq1} 
>>> for key, val in seq2: 
... d.setdefault(key,[]).append(val) 
... 
>>> d 
{1: [[1, 2, 3, 4], [1, 2, 3, 4], [5, 6, 7]], 2: [[5, 6, 7]]} 

您还可以使用collections.defaultdict此类任务。

而且它显示了它的力量，当你有不同的密钥在seq2例如：

>>> seq2 = [[1, [1,2,3,4]], [5, [5,6,7]]] 
>>> d={key: [val] for key, val in seq1} 
>>> for key, val in seq2: 
... d.setdefault(key,[]).append(val) 
... 
>>> d 
{1: [[1, 2, 3, 4], [1, 2, 3, 4]], 2: [[5, 6, 7]], 5: [[5, 6, 7]]} 

如果你不想让你重复可以使用defaultdict与set集装箱：

>>> from collections import defaultdict 
>>> seq1 = [[1, [1,2,3,4]], [2, [5,6,7]]] 
>>> seq2 = [[1, [1,2,3,4]], [1, [5,6,7]]] 
>>> 
>>> d=defaultdict(set) 
>>> for key, val in seq1+seq2: 
... d[key].add(tuple(val)) 
... 
>>> d 
defaultdict(<type 'set'>, {1: set([(5, 6, 7), (1, 2, 3, 4)]), 2: set([(5, 6, 7)])}) 

Answer 3

是的，defaultdict作品也：

from collections import defaultdict 

seq1 = [[1, [1,2,3,4]], [2, [5,6,7]]] 
seq2 = [[1, [1,2,3,4]], [1, [5,6,7]]] 

d={key: [val] for key, val in seq1} 
d = defaultdict(list) 
for key, val in seq2: 
    d[key].append(val) 

print d 

然后：

[(1, [[1, 2, 3, 4], [5, 6, 7]]) 

或者，如果我们删除SEQ1，

seq2 = [[1, [1,2,3,4]], [1, [5,6,7]]] 
d={key: [val] for key, val in seq2} 
d = defaultdict(list) 
for key, val in seq2: 
    d[key].append(val) 
print d 

同样，你将拥有：

defaultdict(<type 'list'>, {1: [[1, 2, 3, 4], [5, 6, 7]]})