Question

我有一个排序的整数ID列表，例如

[1, 2, 10, 15, 16, 17, 20, 34, ...]

在按ID排序的ID旁边，我有一个代码元组（tuple1），

((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"), ...)

我还有另一个元组（tuple2），其格式相同，例如

((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"), ...)

我想将元组组合成一个字典，其中键是ID，值是一个列表，该列表按顺序包含来自tuple1的代码和来自tuple2的代码。如果该ID存在于ID列表中，但不存在于元组中，则该值应为"N/A"。

因此，使用以上数据，应产生以下内容：

{1: ["A", "B"], 2: ["A", "B"], 10: ["N/A", "B"], 15: ["B", "N/A"],
 16: ["A", "A"], 17: ["B", "B"], 20: ["N/A", "N/A"], 34: ["B", "B"]}

我花了很长时间思考这个问题，但我无法提出解决方案。如果有人可以帮助我弄清楚如何在Python中运行它，那将非常有帮助。

谢谢。

编辑：这不是重复的，这个问题要复杂得多。

Answer 1

如果将元组的元组放入字典中，将容易得多。如果缺少密钥，请使用get为字典设置默认值：

ids = [1, 2, 10, 15, 16, 17, 20, 34]
tup1 = ((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"))
tup2 = ((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"))

tup1 = dict(tup1)
tup2 = dict(tup2)    
{k: [tup1.get(k, 'N/A'), tup2.get(k, 'N/A')] for k in ids}

Answer 2

这是对此的一种可能的解决方案：

from collections import defaultdict

keys = [1, 2, 10, 15, 16, 17, 20, 34]

t1 = ((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"))
t2 = ((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"))

merge_d = defaultdict(list)
for d in map(dict, (t1, t2)):
    for k in keys:
        merge_d[k].append(d.get(k, "N/A"))

结果字典将包含：

>>> merge_d
defaultdict(<class 'list'>, {1: ['A', 'B'], 2: ['A', 'B'], 10: ['N/A', 'B'], 15: ['B', 'N/A'], 16: ['A', 'A'], 17: ['B', 'B'], 20: ['N/A', 'N/A'], 34: ['B', 'B']})

Answer 3

有点粗大，但是您无需转换为字典就可以做到：

myList = [1, 2, 10, 15, 16, 17, 20, 34]
tuples1 = ((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"))
tuples2 = ((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"))

myDict = {} 

for i in myList:
    myDict[i] = ["N/A","N/A"] 

for t in tuples1: 
    if(t[0]) in myList:
        for key in myDict:
            if key == t[0]:
                myDict[key][0] = t[1]

for t in tuples2: 
    if(t[0]) in myList:
        for key in myDict:
            if key == t[0]:
                myDict[key][1] = t[1]

print(myDict)

赠予：

{1：['A'，'B']，2：['A'，'B']，10：['N / A'，'B']，15：['B'，' N / A']，16：['A'，'A']，17：['B'，'B']，20：['N / A'，'N / A']，34：[' B'，'B']}

Answer 4

这可以通过另一种方式完成！

lis=[1, 2, 10, 15, 16, 17, 20, 34]
tuple1=((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"))
tuple2=((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"))

z={**dict.fromkeys(lis,'N/A'), **dict(tuple1)}
y={**dict.fromkeys(lis, 'N/A'), **dict(tuple2)}    


ds = [z, y]
d = {}
for k in y.keys():
    d[k] = list(d[k] for d in ds)

Answer 5

我的两分钱：一种单行解决方案，基于@busybear的回答，能够处理任意数量的元组：

ids = [1, 2, 10, 15, 16, 17, 20, 34]
vals = (
    ((1, "A"), (2, "A"), (15, "B"), (16, "A"), (17, "B"), (34, "B"),),
    ((1, "B"), (2, "B"), (10, "B"), (16, "A"), (17, "B"), (34, "B"),),
)

d = {i: [d.get(i, 'N/A') for d in map(dict, vals)] for i in ids}

将两个元组合并到字典中时出现问题

5 个答案: