使用substring打印出多个列表

Question

I have a dictionary with lists and want to create a function that can find lists by using "substring". So that if the list's name is "Things to do on Monday" you only have to type "Monday" to find the list.

Here is what I have so far.

A dictionary.

list_dict = {}
list_dict["Things to do on Monday"] = ['clean', 'workout', 'buy food']
list_dict["Grocery list"] = ['oranges', 'apples', 'milk']
list_dict["Monday outfit"] = ['jeans', 't-shirt', 'shoes']

A function.

def substring_function(substring):
    for key in list_dict.keys():
        if substring in key:
            return list_dict[key]

Then, I do for example:

print(substring_function("Monday"))

The problem is that as you can see I have the word "Monday" in two lists, and I can't figure out how to make it return BOTH lists.

I guess there is something wrong with my for-loop.

Can anyone help me out? Thanks a lot!

Answer 1

You can make the function a generator by replacing the return statement with a yield statement instead:

def substring_function(substring):
    for key in list_dict.keys():
        if substring in key:
            yield list_dict[key]

so that list(substring_function("Monday")) will return a list of matches.

Answer 2

You need add every ocurrency in one var and return after the loop finish

def substring_function(substring):
    rs = list()
    for key in list_dict.keys():
        if substring in key:
            rs.append(list_dict[key])
    return rs

Answer 3

You can try link this.

>>> d = {}
>>> d["Things to do on Monday"] = ['clean', 'workout', 'buy food']
>>> d["Grocery list"] = ['oranges', 'apples', 'milk']
>>> d["Monday outfit"] = ['jeans', 't-shirt', 'shoes']
>>> 
>>> d
{'Things to do on Monday': ['clean', 'workout', 'buy food'], 'Monday outfit': ['jeans', 't-shirt', 'shoes'], 'Grocery list': ['oranges', 'apples', 'milk']}
>>> 

>>> def substring_function(substring, list_dict):
...     final_list = []
...     for key in list_dict:
...         if substring in key:
...             final_list.append(list_dict[key])
...     return final_list
... 
>>> substring_function("Monday", d)
[['clean', 'workout', 'buy food'], ['jeans', 't-shirt', 'shoes']]
>>> 

Further, you can pass the returned object to any function as follows (by destructing it using * ).

>>> def print_items(*args):
...     for arg in args:
...         print(arg)
... 
>>> 
>>> print_items(*substring_function("Monday", d))
['clean', 'workout', 'buy food']
['jeans', 't-shirt', 'shoes']
>>> 
>>> # OR 
... 
>>> items = substring_function("Monday", d)
>>> print_items( *items )
['clean', 'workout', 'buy food']
['jeans', 't-shirt', 'shoes']
>>> 

Answer 4

How about just looking for the 'monday' key(or the key you want to search for):

for key in d.keys():
    if 'monday' in key.lower():
        print(key, d[key])
#Output:
Things to do on Monday ['clean', 'workout', 'buy food']
Monday outfit ['jeans', 't-shirt', 'shoes']

and if you want both lists in one:

def substring(key_to_look_for):
    test = []
    for key in d.keys():
        if key_to_look_for in key.lower():
            test.append(d[key])
    return test

print(substring('monday'))
#Output:
[['clean', 'workout', 'buy food'], ['jeans', 't-shirt', 'shoes']]

With list comprehensions:

def substring(key_to_look_for):
    l = [d[key] for key in d.keys() if key_to_look_for in key.lower()]
    return l
print(substring('monday'))
#Output:
[['clean', 'workout', 'buy food'], ['jeans', 't-shirt', 'shoes']]

List comprehension with lambdas:

def substring(key_to_look_for):
    l = [d[i] for i in list(filter(lambda x: key_to_look_for in x.lower(), d.keys()))]
    return l
substring('monday')
#Output:
[['clean', 'workout', 'buy food'], ['jeans', 't-shirt', 'shoes']]