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经过几个小时和其他帖子的建议，我无法解决我的问题。我必须管理许多字典（直到现在我知道的唯一工作方式）。

对于我要组合的所有四个字典，其中三个具有相同的键（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]}

第四个字典是由包含元数据的参考文件生成的字典，它们的键等于一个值，d1我想要的是创建一个字典，其中包含来自最终字典的信息d1, d2和d3之后的信息。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]}

并以表格格式打印如下：

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“有效”。

#!/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

问题是最后 3 行。

由于缺乏深入的知识，简单的建议（对于傻瓜）将不胜感激。

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1 回答 1

1

我认为这就是您要寻找的东西，尽管关于为什么排除 , , , 等变量的x6逻辑x7尚不y6清楚y7：

首先，使这些变量（例如x1，x2等）存在，并将它们自己的名称作为字符串分配给它们的值，以便以后更容易跟踪结果：

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，让我们实例化您的字典：

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，让我们将字典合并成一个大的 final 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，要从 中获取前两个单独的字母d4，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']}

这基本上是您想要的结果，除了它包括 6 和 7。您能否提供一些背景信息来说明为什么您想要的输出看起来是这样的？无论如何，这应该让你开始。

于 2016-11-21T00:08:28.117 回答