按键和值组合多个字典？

Question

工作时间和其他帖子的建议后，我无法解决我的问题。 我必须管理许多字典（直到现在我知道工作的唯一方式）。

对于我想合并的所有四本字典，它们中的三个具有相同的键（d1，d2和d3）。

d1 = {key1: [x1, x2, x3], key2: [y1, y2, y3], key3: [z1, z2, z3]} 
d2 = {key1: [x4, x5, x6],key2: [y4, y5, y6], key3: [z4, z5, z6]} 
d3 = {key1: [x7, x8, x9], key2: [y7, y8, y9], key3: [z7, z8, z9]} 
d4 = {x2: [a, b, c], y2: [d, e, f], z2: [g, h, i]} 

第四词典是由包含元数据的引用文件中生成一个字典，它们的键在d1等于一个值和我想是创建具有从d1, d2和d3的信息的字典，并包括后最终字典中的信息d4。

final_dict = {key1: [x1, a, b, x2, x3, x4, x5, x8, x9], 
       key2: [y1, d, e, y2, y3, y4, y5, y8, y9], 
       key3: [z1, g, h, z2, z3, z4, z5, z8, z9]} 

，并以表格的形式打印这样的：

key1 x1 a b x2 x3 x4 x5 x8 x9 
key2 y1 d e y2 y3 y4 y5 y8 y9 
key3 z1 g h z2 z3 z4 z5 z8 z9 

目前，我有一个肮脏的脚本，但“作品”。

#!/usr/bin/env python 

with open("file1.txt", "r") as file1, open("file2.txt", "r") as file2,/
    open("file3.txt", "r") as file3, open("file4.txt", "r") as file4: 

    d1 = {} 
    d2 = {} 
    d3 = {} 
    d4 = {} 
    dicts = [d1, d2, d3, d4] 

    #d1 = {key1: [x1, x2, x3], key2: [y1, y2, y3], key3: [z1, z2, z3]} 
    #d2 = {key1: [x4, x5, x6],key2: [y4, y5, y6], key3: [z4, z5, z6]} 
    #d3 = {key1: [x7, x8, x9], key2: [y7, y8, y9], key3: [z7, z8, z9]} 
    #d4 = {x2: [a, b, c], y2: [d, e, f], z2: [g, h, i]} 

    for b in file1: 
     row = b.strip().split('\t') 
     if row[0] not in d1: 
      d1[row[0]] = row[1], row[3], row[4] 

    for c in file2: 
     row = c.strip().split('\t') 
     if row[0] not in d2: 
      d2[row[0]] = row[1:] 

    for f in file3: 
     row = f.strip().split('\t') 
     if row[0] not in d3: 
      d3[row[0]] = row[1:] 

    for m in file4: 
     row = m.strip().split('\t') 
     if row[0] not in d4: 
      d4[row[0]] = row[1], row[3], row[2] 

    final_dict = {} 
    for k in (dicts): 
     for key, value in k.iteritems(): 
      final_dict[key].append(value) 

    print final_dic 

    #key1 x1 a b x2 x3 x4 x5 x8 x9 
    #key2 y1 d e y2 y3 y4 y5 y8 y9 
    #key3 z1 g h z2 z3 z4 z5 z8 z9 

的问题是最后的3行。

由于缺乏深厚的知识，我们将不胜感激简单的建议（傻瓜）。

Answer 1

我觉得这是你要找的东西，虽然逻辑，为什么变量像x6，x7，y6，y7等被排除在外是目前还不清楚：

首先，这些变量（例如x1，x2等）的存在，并指定自己的名字作为字符串，其值后的结果更容易追踪：

values = [letter + str(number) for letter in 'xyz' for number in range(1, 10)] + ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i'] 
for v in values: 
    exec('%s = "%s"' % (v, v)) 

接下来，让我们来实例化你的字典：

d1 = {key1: [x1, x2, x3], key2: [y1, y2, y3], key3: [z1, z2, z3]} 
d2 = {key1: [x4, x5, x6],key2: [y4, y5, y6], key3: [z4, z5, z6]} 
d3 = {key1: [x7, x8, x9], key2: [y7, y8, y9], key3: [z7, z8, z9]} 
d4 = {x2: [a, b, c], y2: [d, e, f], z2: [g, h, i]} 

然后，让我们合并的字典为一体，大，最终dict：

new_dict = {} 
for d in [d1, d2, d3]: 
    for key in d: 
     if key not in new_dict: 
      # if key not yet in the dict, make it so 
      new_dict[key] = d[key] 
     else: 
      # if key already there, then we'll just add the lists together 
      new_dict[key] += d[key] 

最后，从d4获得前两个单个字母，我们可以试试这个：

for key in new_dict: 
    for other_key in d4: 
     if other_key in new_dict[key]: 
      new_dict[key] += d4[other_key][:2] 

检查输出：

>>> new_dict 
{'key2': ['y1', 'y2', 'y3', 'y4', 'y5', 'y6', 'y7', 'y8', 'y9', 'd', 'e'], 'key3': ['z1', 'z2', 'z3', 'z4', 'z5', 'z6', 'z7', 'z8', 'z9', 'g', 'h'], 'key1': ['x1', 'x2', 'x3', 'x4', 'x5', 'x6', 'x7', 'x8', 'x9', 'a', 'b']} 

这基本上是你想要的结果，除了它包括6和7。你能提供一些背景知道为什么你想要的输出看起来像这样吗？无论如何，这应该让你开始。