# Finding and replacing elements in a list

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I have to search through a list and replace all occurrences of one element with another. So far my attempts in code are getting me nowhere, what is the best way to do this?

For example, suppose my list has the following integers

``````>>> a = [1,2,3,4,5,1,2,3,4,5,1]
``````

and I need to replace all occurrences of the number 1 with the value 10 so the output I need is

``````>>> a = [10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]
``````

Thus my goal is to replace all instances of the number 1 with the number 10.

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## Finding and replacing elements in a list find: Questions

Finding the index of an item in a list

5 answers

Given a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, how do I get its index (`1`) in Python?

3740

Answer #1

``````>>> ["foo", "bar", "baz"].index("bar")
1
``````

Reference: Data Structures > More on Lists

# Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, `index` is a rather weak component of the `list` API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about `list.index` follow. It is probably worth initially taking a look at the documentation for it:

``````list.index(x[, start[, end]])
``````

Return zero-based index in the list of the first item whose value is equal to x. Raises a `ValueError` if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

## Linear time-complexity in list length

An `index` call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give `index` a hint. For instance, in this snippet, `l.index(999_999, 999_990, 1_000_000)` is roughly five orders of magnitude faster than straight `l.index(999_999)`, because the former only has to search 10 entries, while the latter searches a million:

``````>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514

``````

## Only returns the index of the first match to its argument

A call to `index` searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

``````>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
``````

Most places where I once would have used `index`, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for `index`, take a look at these excellent Python features.

## Throws if element not present in list

A call to `index` results in a `ValueError` if the item"s not present.

``````>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
``````

If the item might not be present in the list, you should either

1. Check for it first with `item in my_list` (clean, readable approach), or
2. Wrap the `index` call in a `try/except` block which catches `ValueError` (probably faster, at least when the list to search is long, and the item is usually present.)

3740

Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

``````>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
...

|
|  index(...)
|      L.index(value, [start, [stop]]) -> integer -- return first index of value
|
``````

which will often lead you to the method you are looking for.

3740

Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use `enumerate()`:

``````for i, j in enumerate(["foo", "bar", "baz"]):
if j == "bar":
print(i)
``````

The `index()` function only returns the first occurrence, while `enumerate()` returns all occurrences.

As a list comprehension:

``````[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
``````

Here"s also another small solution with `itertools.count()` (which is pretty much the same approach as enumerate):

``````from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
``````

This is more efficient for larger lists than using `enumerate()`:

``````\$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
\$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop
``````

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