如何提取访问列表指定子列表索引值？含匹配及字典场景需求

Got it, let's break this down for both your nested list and dictionary scenarios—no messy nested loops required, just clean, readable one-liners as you requested:

First, let's handle your existing list structure [[object1Name, string_1, string_2], ...]:

Check if myString exists in any string_1 position #

You can use a generator expression inside the in operator to check this in one line:

myString in (item[1] for item in my_list)

This iterates through each sublist, grabs the second element (index 1), and checks if myString matches any of them—way cleaner than nested loops.

Get indices of matching items #

If you need all indices where the sublist's string_1 equals myString, use a list comprehension:

matching_indices = [idx for idx, item in enumerate(my_list) if item[1] == myString]

If you only need the first matching index (and want a default value like None if there's no match), use next() with a generator:

first_match_idx = next((idx for idx, item in enumerate(my_list) if item[1] == myString), None)

Now let's assume your dictionary maps object names to their string pairs (adjust if your actual structure differs):

my_dict = {
    "object1Name": {"string_1": "s1_val", "string_2": "s2_val"},
    "object2Name": {"string_1": "s3_val", "string_2": "s4_val"},
    # ... more entries
}

Check if myString exists in any string_1 value #

Again, use a generator with in for a clean one-liner:

myString in (entry["string_1"] for entry in my_dict.values())

Get matching keys (dictionary equivalent of indices) #

Since dictionaries use keys instead of numeric indices, to get all keys where the string_1 value matches:

matching_keys = [key for key, entry in my_dict.items() if entry["string_1"] == myString]

For just the first matching key (with a default None if no match):

first_match_key = next((key for key, entry in my_dict.items() if entry["string_1"] == myString), None)

If your dictionary has a different structure (e.g., keys are numeric indices and values are the original sublists), just tweak the accessors—the core logic stays the same!

内容的提问来源于stack exchange，提问作者Nec Xelos