从两个不同长度的列表中添加到字典中的值

Question

我有四个列表,如下所示:

lista = [['l', 'k'],['e', '3'],['c', 'k'],['x', 'i'],['d', 'f']]
listanum = [1,2,3,4,5]
listb = [['a', 'k'],['c', 'm'],['v', 'f']]
listbnum = [1,3,4]

lista并且listanum是同步的,listb和listbnum也.我想要一本字典,其中键都在项目listanum和值在项目lista和listb,结果将是:

di = {1: [['l','k'],['a', 'k']], 
      2: [['e', '3'],[]], 
      3:[['c','k'],['c', 'm']], 
      4: [['x', 'i'],['v', 'f']], 
      5: [['d', 'f'][]]
}

因此,如果listb中listanum中的数字没有值,则dictionarys值中的第二个列表为空.

我试过这个:

di = {}
for i in xrange(len(lista)):
  pos = listanum[i]
  if pos not in di:
    di[pos] = [[],[]]
  di[pos][0].append(lista[i])  
  if i in listbnum:
    di[pos][1].append(listb[i])

但得到此错误消息:'IndexError: list index out of range'.我不明白为什么它超出范围???

Answer 1

In [7]: da = dict(zip(listanum, lista))

In [8]: db = dict(zip(listbnum, listb))

In [9]: {k:[da.get(k,[]), db.get(k,[])] for k in set(listanum + listbnum)}
Out[9]: 
{1: [['l', 'k'], ['a', 'k']],
 2: [['e', '3'], []],
 3: [['c', 'k'], ['c', 'm']],
 4: [['x', 'i'], ['v', 'f']],
 5: [['d', 'f'], []]}

Answer 2

压缩项目,并使用collections.defaultdict默认值为lists:

from itertools import chain
from collections import defaultdict

di = defaultdict(list)

for key, value in chain(zip(listanum, lista), zip(listbnum, listb)):
    di[key].append(value)

我曾经chain更容易在两组键值对上进行循环; 这适用于Python 2和3.如果这只是Python 2代码,您可以使用+连接两个列表.

输出pprint并转换回常规dict以使打印更容易:

>>> pprint(dict(di))
{1: [['l', 'k'], ['a', 'k']],
 2: [['e', '3']],
 3: [['c', 'k'], ['c', 'm']],
 4: [['x', 'i'], ['v', 'f']],
 5: [['d', 'f']]}

这不会为第二组创建空列表; 如果你必须有空列表,你只能构建两个单独的词典然后合并它们:

dicta = dict(zip(listanum, lista))
dictb = dict(zip(listbnum, listb))

di = {k: [dicta.get(k, []), dictb.get(k, [])] for k in dicta.viewkeys() | dictb.viewkeys()}

对于Python 2,使用Python 3 .keys()来代替.viewkeys(),以产生:

>>> pprint(di)
{1: [['l', 'k'], ['a', 'k']],
 2: [['e', '3'], []],
 3: [['c', 'k'], ['c', 'm']],
 4: [['x', 'i'], ['v', 'f']],
 5: [['d', 'f'], []]}

具体来说,对于您的代码,您将i(索引lista)与pos以下内容混淆:

  if i in listbnum:
    di[pos][1].append(listb[i])

对于i = 4,i in listbnum是True,但listb[4]不存在.您的代码也尝试附加来自lista和的列表listb,这不会导致正确的输出.

使用/ lists 的单独循环更改您的版本以使其工作:listblistbnum

di = {}
for i, pos in enumerate(listanum):
    if pos not in di:
        di[pos] = [[],[]]
    di[pos][0][:] = lista[i]

for i, pos in enumerate(listbnum):
    di[pos][1][:] = listb[i]