Unlock Python List Index: Fast Lookup, Pitfalls, and Pro Tips

Problem: Given a list ["年薪10万", "年薪30万", "年薪50万", "年薪100万"], find the index of "年薪100万".

The simplest solution uses the list index method:

["年薪10万", "年薪30万", "年薪50万", "年薪100万"].index("年薪100万")

3

Although concise, list.index is relatively weak and can have performance issues on large lists.

Full signature: list.index(x[, start[, end]]) returns the first index of x or raises ValueError if not found. Optional start and end limit the search range but the returned index is relative to the whole list.

Each call scans elements sequentially until a match is found, which can become a bottleneck for long lists when the element’s position is unknown.

Providing a narrower search range can dramatically speed up the lookup, e.g., l.index(999_999, 999_990, 1_000_000) can be up to ten thousand times faster than l.index(999_999) on a list of one million items.

# timing example
import timeit
timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
# → 15.676...
timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
# → 0.000329...

The index method returns only the first matching index. To obtain all matches, use a list comprehension or generator expression:

[i for i, e in enumerate([1, 2, 1]) if e == 1]  # → [0, 2]

If the element is absent, index raises ValueError. Wrap the call in try/except to handle this case:

def find_element_in_list(element, lst):
    try:
        return lst.index(element)
    except ValueError:
        return 'not exists'

Tip: Use the built‑in help function on a list object to view all its methods, including index:

help(["年薪10万", "年薪30万", "年薪50万", "年薪100万"])