Python - 根据前者的键值将字典的键附加为另一个键中的键值

Question

I have a dictionary that is one where each key is associated to a list. For example:

a = ['apple', 'pear', 'banana']
b = ['car', 'plane', 'boat']
c = ['cat', 'dog', 'bird']
d = {'a' : a, 'b' : b, 'c' : c}

And I have a list of dictionaries where a particular key value in each dictionary will be a string that matches one of the strings from the lists. Such as:

x = [{'thing' : 'apple'}, {'thing': 'dog'}, {'thing' : 'boat'}]

What I want to do is add a key to each dictionary where the value matches the name of the list where the string is found. Resulting in:

x = [{'thing' : 'apple', 'list' : 'a'}, {'thing' : 'dog', 'list' : 'c']}, {'thing': 'boat', 'list': 'b'}]

I have tried

for k in d:
    for m in x:
        if m['thing'] in k:
            m['list'] = k

I have a feeling that I'm overcomplicating this but haven't been able to figure out where I'm going wrong. Any advice is appreciated.

Edit: Something I forgot to mention in my post when translating it to more general terms is the that the strings which are in a,b, and c are substrings of those found in x. So x would actually be more like x = [{'thing' : 'apple | fruit'}, {'thing': 'dog | animal'}, {'thing' : 'boat | vehicle'}]

Answer 1

Check this code

a = ['apple', 'pear', 'banana']
b = ['car', 'plane', 'boat']
c = ['cat', 'dog', 'bird']
d = {'a' : a, 'b' : b, 'c' : c}

x = [{'thing' : 'apple'}, {'thing': 'dog'}, {'thing' : 'boat'}]

# with  list comprehension
nx = [{'thing' : m['thing'], 'list' : key} for key, listVals in d.items() for m in x if m['thing'] in listVals]

# normal way.
# nx = []
# for m in x:
#     for key, listVals in d.items():
#         if m['thing'] in listVals :
#             nx.append({'thing' : m['thing'], 'list' : key})

print(nx)

Answer 2

One option is to create an inverse of d where the keys are the values in each of the lists and the values are the keys in d . Then use this inverse dictionary to update x .

First create the inverse dict:

d_inv = {d[k][i]: k for k in d for i in range(len(d[k]))}
print(d_inv)
#{'apple': 'a', 'banana': 'a', 'car': 'b', 'pear': 'a', 'dog': 'c', 'cat': 'c', 
# 'plane': 'b', 'bird': 'c', 'boat': 'b'}

This assumes that you do not have the same element appearing in more than one list.

Now update x :

for elem in x:
    elem['list'] = d_inv[elem['thing']]
print(x)
[
    {'thing': 'apple', 'list': 'a'},
    {'thing': 'dog', 'list': 'c'},
    {'thing': 'boat', 'list': 'b'}
]

Answer 3

I appreciate the above comment(s) for this problem.

As we are searching for a key inside list and a , b , c are keys of dictionary that refers/points to list. k in if m['thing'] in k refers key not the list so it should be changed to d[k] .

http://rextester.com/JVRG57284

Have a look at your modified code below (please comment if you're not satisfied with the code or if it doesn't satisfy your need or if it fail for any of your test cases):

import json

a = ['apple', 'pear', 'banana']
b = ['car', 'plane', 'boat']
c = ['cat', 'dog', 'bird']
d = {'a' : a, 'b' : b, 'c' : c}

x = [{'thing' : 'apple'}, {'thing': 'dog'}, {'thing' : 'boat'}]

for k in d:
    for m in x:
        if m['thing'] in d[k]:
            m['list'] = k

# pretty printing x dictionary
print(json.dumps(x, indent=4))

"""
[
    {
        "list": "a",
        "thing": "apple"
    },
    {
        "list": "c",
        "thing": "dog"
    },
    {
        "list": "b",
        "thing": "boat"
    }
]
"""

Answer 4

You can use a list comprehension and dictionary unpacking:

a = ['apple', 'pear', 'banana']
b = ['car', 'plane', 'boat']
c = ['cat', 'dog', 'bird']
d = {'a' : a, 'b' : b, 'c' : c}
x = [{'thing' : 'apple'}, {'thing': 'dog'}, {'thing' : 'boat'}]
final_results = [{**i, **{'list':[a for a, b in d.items() if i['thing'] in b][0]}} for i in x]

Output:

 [{'list': 'a', 'thing': 'apple'}, {'list': 'c', 'thing': 'dog'}, {'list': 'b', 'thing': 'boat'}]