如何在字典列表中引用和返回值

Question

I have a list of lists.

Inside of each list there are a few thousand lists of dictionaries. One list might contain multiple dictionaries, one dictionary, or it might be empty.

Here is an abridged list with three examples rows in the list:

list_of_lists = [[], [{'text': 'analytics', 'indices': [18, 28]}, {'text': 'datascience', 'indices': [35, 47]}, {'text': 'restaurants', 'indices': [54, 66]}, {'text': 'machinelearning', 'indices': [92, 108]}, {'text': 'bigData', 'indices': [109, 117]}, {'text': 'CRM', 'indices': [118, 122]}], [{'text': 'python', 'indices': [49, 56]}, {'text': 'datascience', 'indices': [57, 69]}]

So within this list there's one empty list, one list with 6 dictionaries, and one list with 2 dictionaries.

I need to extract the value from the key:value pair that includes 'text': 'SOME_STRING'.

Also, IMPORTANTLY, each value should return in a list with the same index from the original input list. In other words, for example, for the second list of 6 key:value pairs, all 6 values should be returned in a list at the same index that it was at in the original list_of_lists

So here is my desired output from the above example:

list_of_values = [[], ['analytics', 'datascience', 'restaurants', 'machinelearning', 'bigData', 'CRM', 'python'], ['python', 'datascience']]

I have written the code below that almost does what I want. It returns a list of all of these strings, but it does not return them at the same index, and it also returns the indices dictionary that I don't want.

new_list_of_value_lists = []
for line in list_of_lists:
    for dictionary in line:
        for key, value in dictionary.items():
            new_list_of_value_lists.append(value)

Answer 1

Create a different list for each nested list of dicts and append to the parent list. The empty list gets zero iterations so the resulting list stays empty, while the others have their values collected in the list comprehension :

list_of_values = []
for lst in list_of_lists:
    list_of_values.append([dct['text'] for dct in lst])

print(list_of_values)
# [[], ['analytics', 'datascience', 'restaurants', 'machinelearning', 'bigData', 'CRM'], ['python', 'datascience']]