# Finding the index of an item in a list

## Question:

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

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

See the documentation for the built-in `.index()` method of the list:

``````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.

## Caveats

### Linear time-complexity in list length

An `index` call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.

This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the `start` and `end` parameters can be used to narrow the search.

For example:

``````>>> 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
``````

The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.

### Only the index of the first match is returned

A call to `index` searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, `index` cannot solve the problem:

``````>>> [1, 1].index(1) # the `1` index is not found.
0
``````
``````>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2
``````

The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.

### Raises an exception if there is no match

As noted in the documentation above, using `.index` will raise an exception if the searched-for value is not in the list:

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

If this is a concern, either explicitly check first using `item in my_list`, or handle the exception with `try`/`except` as appropriate.

The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.

`index()` returns the first index of value!

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

``````def all_indices(value, qlist):
indices = []
idx = -1
while True:
try:
idx = qlist.index(value, idx+1)
indices.append(idx)
except ValueError:
break
return indices

all_indices("foo", ["foo","bar","baz","foo"])
``````
``````a = ["foo","bar","baz",'bar','any','much']

indexes = [index for index in range(len(a)) if a[index] == 'bar']
``````

A problem will arise if the element is not in the list. This function handles the issue:

``````# if element is found it returns index of element else returns None

def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
``````

All of the proposed functions here reproduce inherent language behavior but obscure what’s going on.

``````[i for i in range(len(mylist)) if mylist[i]==myterm]  # get the indices

[each for each in mylist if each==myterm]             # get the items

mylist.index(myterm) if myterm in mylist else None    # get the first index and fail quietly
``````

Why write a function with exception handling if the language provides the methods to do what you want itself?

Simply you can go with

``````a = [['hand', 'head'], ['phone', 'wallet'], ['lost', 'stock']]
b = ['phone', 'lost']

res = [[x[0] for x in a].index(y) for y in b]
``````

Another option

``````>>> a = ['red', 'blue', 'green', 'red']
>>> b = 'red'
>>> offset = 0;
>>> indices = list()
>>> for i in range(a.count(b)):
...     indices.append(a.index(b,offset))
...     offset = indices[-1]+1
...
>>> indices
[0, 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
``````

To get all indexes:

``````indexes = [i for i, x in enumerate(xs) if x == 'foo']
``````

A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:

``````>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
``````

You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.

You have to set a condition to check if the element you’re searching is in the list

``````if 'your_element' in mylist:
print mylist.index('your_element')
else:
print None
``````

# And now, for something completely different…

… like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices — even if it is an empty list. It works with strings as well.

``````def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval

l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
``````

When pasted into an interactive python window:

``````Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
>>> def indices(the_list, val):
...     """Always returns a list containing the indices of val in the_list"""
...     retval = []
...     last = 0
...     while val in the_list[last:]:
...             i = the_list[last:].index(val)
...             retval.append(last + i)
...             last += i + 1
...     return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
``````

# Update

After another year of heads-down python development, I’m a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.

Something like this:

``````def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]

l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
``````

Which, when pasted into an interactive python window yields:

``````Python 2.7.14 |Anaconda, Inc.| (default, Dec  7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
>>> def indices(l, val):
...     """Always returns a list containing the indices of val in the_list"""
...     return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
``````

And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn’t even see that answer, because I didn’t understand it. I hope that my somewhat more verbose example will aid understanding.

If the single line of code above still doesn’t make sense to you, I highly recommend you Google ‘python list comprehension’ and take a few minutes to familiarize yourself. It’s just one of the many powerful features that make it a joy to use Python to develop code.

This solution is not as powerful as others, but if you’re a beginner and only know about `for`loops it’s still possible to find the first index of an item while avoiding the ValueError:

``````def find_element(p,t):
i = 0
for e in p:
if e == t:
return i
else:
i +=1
return -1
``````
``````name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
new_list=[]
for item in list:
new_list.append(item[0])
print(new_list)
try:
location= new_list.index(name)
except:
location=-1
print (location)
``````

This accounts for if the string is not in the list too, if it isn’t in the list then `location = -1`

All indexes with the `zip` function:

``````get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]

print get_indexes(2, [1, 2, 3, 4, 5, 6, 3, 2, 3, 2])
print get_indexes('f', 'xsfhhttytffsafweef')
``````

If you want all indexes, then you can use NumPy:

``````import numpy as np

array = [1, 2, 1, 3, 4, 5, 1]
item = 1
np_array = np.array(array)
item_index = np.where(np_array==item)
print item_index
# Out: (array([0, 2, 6], dtype=int64),)
``````

### Getting all the occurrences and the position of one or more (identical) items in a list

With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.

``````>>> alist = ['foo', 'spam', 'egg', 'foo']
>>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo']
>>> foo_indexes
[0, 3]
>>>
``````

### Let’s make our function findindex

This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.

``````def indexlist(item2find, list_or_string):
"Returns all indexes of an item in a list or a string"
return [n for n,item in enumerate(list_or_string) if item==item2find]

print(indexlist("1", "010101010"))
``````

Output

``````[1, 3, 5, 7]
``````

## Simple

``````for n, i in enumerate([1, 2, 3, 4, 1]):
if i == 1:
print(n)
``````

Output:

``````0
4
``````

Since Python lists are zero-based, we can use the zip built-in function as follows:

``````>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]
``````

where “haystack” is the list in question and “needle” is the item to look for.

(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)

## Finding the index of an item given a list containing it in Python

For a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, what’s the cleanest way to get its index (1) in Python?

Well, sure, there’s the index method, which returns the index of the first occurrence:

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

There are a couple of issues with this method:

• if the value isn’t in the list, you’ll get a `ValueError`
• if more than one of the value is in the list, you only get the index for the first one

### No values

If the value could be missing, you need to catch the `ValueError`.

You can do so with a reusable definition like this:

``````def index(a_list, value):
try:
return a_list.index(value)
except ValueError:
return None
``````

And use it like this:

``````>>> print(index(l, 'quux'))
None
>>> print(index(l, 'bar'))
1
``````

And the downside of this is that you will probably have a check for if the returned value `is` or `is not` None:

``````result = index(a_list, value)
if result is not None:
do_something(result)
``````

### More than one value in the list

If you could have more occurrences, you’ll not get complete information with `list.index`:

``````>>> l.append('bar')
>>> l
['foo', 'bar', 'baz', 'bar']
>>> l.index('bar')              # nothing at index 3?
1
``````

You might enumerate into a list comprehension the indexes:

``````>>> [index for index, v in enumerate(l) if v == 'bar']
[1, 3]
>>> [index for index, v in enumerate(l) if v == 'boink']
[]
``````

If you have no occurrences, you can check for that with boolean check of the result, or just do nothing if you loop over the results:

``````indexes = [index for index, v in enumerate(l) if v == 'boink']
for index in indexes:
do_something(index)
``````

### Better data munging with pandas

If you have pandas, you can easily get this information with a Series object:

``````>>> import pandas as pd
>>> series = pd.Series(l)
>>> series
0    foo
1    bar
2    baz
3    bar
dtype: object
``````

A comparison check will return a series of booleans:

``````>>> series == 'bar'
0    False
1     True
2    False
3     True
dtype: bool
``````

Pass that series of booleans to the series via subscript notation, and you get just the matching members:

``````>>> series[series == 'bar']
1    bar
3    bar
dtype: object
``````

If you want just the indexes, the index attribute returns a series of integers:

``````>>> series[series == 'bar'].index
Int64Index([1, 3], dtype='int64')
``````

And if you want them in a list or tuple, just pass them to the constructor:

``````>>> list(series[series == 'bar'].index)
[1, 3]
``````

Yes, you could use a list comprehension with enumerate too, but that’s just not as elegant, in my opinion – you’re doing tests for equality in Python, instead of letting builtin code written in C handle it:

``````>>> [i for i, value in enumerate(l) if value == 'bar']
[1, 3]
``````

## Is this an XY problem?

Why do you think you need the index given an element in a list?

If you already know the value, why do you care where it is in a list?

If the value isn’t there, catching the `ValueError` is rather verbose – and I prefer to avoid that.

I’m usually iterating over the list anyways, so I’ll usually keep a pointer to any interesting information, getting the index with enumerate.

If you’re munging data, you should probably be using pandas – which has far more elegant tools than the pure Python workarounds I’ve shown.

I do not recall needing `list.index`, myself. However, I have looked through the Python standard library, and I see some excellent uses for it.

There are many, many uses for it in `idlelib`, for GUI and text parsing.

The `keyword` module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming.

In Lib/mailbox.py it seems to be using it like an ordered mapping:

``````key_list[key_list.index(old)] = new
``````

and

``````del key_list[key_list.index(key)]
``````

In Lib/http/cookiejar.py, seems to be used to get the next month:

``````mon = MONTHS_LOWER.index(mon.lower())+1
``````

In Lib/tarfile.py similar to distutils to get a slice up to an item:

``````members = members[:members.index(tarinfo)]
``````

In Lib/pickletools.py:

``````numtopop = before.index(markobject)
``````

What these usages seem to have in common is that they seem to operate on lists of constrained sizes (important because of O(n) lookup time for `list.index`), and they’re mostly used in parsing (and UI in the case of Idle).

While there are use-cases for it, they are fairly uncommon. If you find yourself looking for this answer, ask yourself if what you’re doing is the most direct usage of the tools provided by the language for your use-case.

For those coming from another language like me, maybe with a simple loop it’s easier to understand and use it:

``````mylist = ["foo", "bar", "baz", "bar"]
newlist = enumerate(mylist)
for index, item in newlist:
if item == "bar":
print(index, item)
``````

I am thankful for So what exactly does enumerate do?. That helped me to understand.

Python `index()` method throws an error if the item was not found. So instead you can make it similar to the `indexOf()` function of JavaScript which returns `-1` if the item was not found:

``````try:
index = array.index('search_keyword')
except ValueError:
index = -1
``````

There is a more functional answer to this.

``````list(filter(lambda x: x[1]=="bar",enumerate(["foo", "bar", "baz", "bar", "baz", "bar", "a", "b", "c"])))
``````

More generic form:

``````def get_index_of(lst, element):
return list(map(lambda x: x[0],
(list(filter(lambda x: x[1]==element, enumerate(lst))))))
``````

Finding index of item x in list L:

``````idx = L.index(x) if (x in L) else -1
``````

# If performance is of concern:

It is mentioned in numerous answers that the built-in method of `list.index(item)` method is an O(n) algorithm. It is fine if you need to perform this once. But if you need to access the indices of elements a number of times, it makes more sense to first create a dictionary (O(n)) of item-index pairs, and then access the index at O(1) every time you need it.

If you are sure that the items in your list are never repeated, you can easily:

``````myList = ["foo", "bar", "baz"]

# Create the dictionary
myDict = dict((e,i) for i,e in enumerate(myList))

# Lookup
myDict["bar"] # Returns 1
# myDict.get("blah") if you don't want an error to be raised if element not found.
``````

If you may have duplicate elements, and need to return all of their indices:

``````from collections import defaultdict as dd
myList = ["foo", "bar", "bar", "baz", "foo"]

# Create the dictionary
myDict = dd(list)
for i,e in enumerate(myList):
myDict[e].append(i)

# Lookup
myDict["foo"] # Returns [0, 4]
``````

As indicated by @TerryA, many answers discuss how to find one index.

`more_itertools` is a third-party library with tools to locate multiple indices within an iterable.

Given

``````import more_itertools as mit

iterable = ["foo", "bar", "baz", "ham", "foo", "bar", "baz"]
``````

Code

Find indices of multiple observations:

``````list(mit.locate(iterable, lambda x: x == "bar"))
# [1, 5]
``````

Test multiple items:

``````list(mit.locate(iterable, lambda x: x in {"bar", "ham"}))
# [1, 3, 5]
``````

See also more options with `more_itertools.locate`. Install via `> pip install more_itertools`.

Letâ€™s give the name `lst` to the list that you have. One can convert the list `lst` to a `numpy array`. And, then use numpy.where to get the index of the chosen item in the list. Following is the way in which you will implement it.

``````import numpy as np

lst = ["foo", "bar", "baz"]  #lst: : 'list' data type
print np.where( np.array(lst) == 'bar')[0][0]

>>> 1
``````

using dictionary , where process the list first and then add the index to it

``````from collections import defaultdict

index_dict = defaultdict(list)
word_list =  ['foo','bar','baz','bar','any', 'foo', 'much']

for word_index in range(len(word_list)) :
index_dict[word_list[word_index]].append(word_index)

word_index_to_find = 'foo'
print(index_dict[word_index_to_find])

# output :  [0, 5]
``````

If you are going to find an index once then using “index” method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect.
Using bisect module on my machine is about 20 times faster than using index method.

Here is an example of code using Python 3.8 and above syntax:

``````import bisect
from timeit import timeit

def bisect_search(container, value):
return (
index
if (index := bisect.bisect_left(container, value)) < len(container)
and container[index] == value else -1
)

data = list(range(1000))
# value to search
value = 666

# times to test
ttt = 1000

t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)

print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")
``````

Output:

``````t1=0.0400, t2=0.0020, diffs t1/t2=19.60
``````

### For one comparable

``````# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2
``````

### Custom predicate

``````some_list = [item1, item2, item3]

# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
i for i, item in enumerate(some_list)
if item.matches_your_criteria())
``````

### Finding index of all items by predicate

``````index_of_staff_members = [
i for i, user in enumerate(users)
if user.is_staff()]
``````

There is a chance that that value may not be present so to avoid this ValueError, we can check if that actually exists in the list .

``````list =  ["foo", "bar", "baz"]

item_to_find = "foo"

if item_to_find in list:
index = list.index(item_to_find)
print("Index of the item is " + str(index))
else:
print("That word does not exist")
``````

It just uses the python function `array.index()` and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function `indexOf()`).

``````fruits = ['apple', 'banana', 'cherry']

try:
pos = fruits.index("mango")
except:
pos = -1
``````

In this case "mango" is not present in the list `fruits` so the `pos` variable is -1, if I had searched for "cherry" the `pos` variable would be 2.

Simple option:

``````a = ["foo", "bar", "baz"]
[i for i in range(len(a)) if a[i].find("bar") != -1]
``````

I find this two solution is better and I tried it by myself

``````>>> expences = [2200, 2350, 2600, 2130, 2190]
>>> 2000 in expences
False
>>> expences.index(2200)
0
>>> expences.index(2350)
1
>>> index = expences.index(2350)
>>> expences[index]
2350

>>> try:
...     print(expences.index(2100))
... except ValueError as e:
...     print(e)
...
2100 is not in list
>>>

``````

Pythonic way would to use `enumerate` but you can also use `indexOf` from `operator` module. Please note that this will raise `ValueError` if `b` not in `a`.

``````>>> from operator import indexOf
>>>
>>>
>>> help(indexOf)
Help on built-in function indexOf in module _operator:

indexOf(a, b, /)
Return the first index of b in a.

>>>
>>>
>>> indexOf(("foo", "bar", "baz"), "bar") # with tuple
1
>>> indexOf(["foo", "bar", "baz"], "bar") # with list
1
``````

Certain structures in python contains a index method that works beautifully to solve this question.

``````'oi tchau'.index('oi')     # 0
['oi','tchau'].index('oi') # 0
('oi','tchau').index('oi') # 0
``````

References:

In lists

In tuples

In string

``````text = ["foo", "bar", "baz"]
target = "bar"

[index for index, value in enumerate(text) if value == target]
``````

For a small list of elements, this would work fine. However, if the
list contains a large number of elements, better to apply binary
search with O(log n) runtime complexity

.

Here’s a two-liner using Python’s `index()` function:

``````LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
``````

Output: `1`

List comprehension would be the best option to acquire a compact implementation in finding the index of an item in a list.

``````a_list = ["a", "b", "a"]
print([index for (index , item) in enumerate(a_list) if item == "a"])
``````
``````me = ["foo", "bar", "baz"]
me.index("bar")
``````

You can apply this for any member of the list to get their index

One can use zip() function to get the index of the value in the list. The code could be;

``````list1 = ["foo","bar","baz"]
for index,value in zip(range(0,len(list1)),list1):
if value == "bar":
print(index)
``````

Try the following code:

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

Don’t. If you absolutely need to, use the `.index(item...)` method on list. However, it takes linear time, and if you find yourself reaching for it, you are probably misusing lists to do something you should not do with them.

Most likely, you care either about 1) a two-way mapping between integers and items, or 2) about finding an item in a sorted list of items.

For the first one, use a pair of dictionaries. If you want a library that does this for you, use the `bidict` library.

For the second one, use methods that can properly make use of the fact that the list is sorted. Use the built-in `bisect` module in python.

If you find yourself wanting to insert items in a sorted list, you should also not use a sorted list. Either weaken the sorted requirement to a heap using the builtin `heapq` module, or use the `sortedcontainers` library.

It is bad practice to use a data structure that is not designed for what you want to do. Using one that matches the task you give it will both telegraph to the reader that you want to do that specific thing, and also make your solution a lot faster/more scalable in practice.

A simple solution in python:

``````li1=["foo","bar","baz"]
for i in range(len(li1)):
if li1[i]=="bar":
print(i)
``````

The data type of the list elements is irrelevant. Just replace the "bar" with the element you are looking for. We can also write a function for this as:

``````def indexfinder(element,lis):
for i in range(len(lis)):
if lis[i]==element:
return i
``````

I found a super handy package for this Huntela, and I’ve been using it ever since.

It has a straightforward interface too:

``````>>> import huntela
>>>
>>> huntela.simple_search(term='bar', items=['foo', 'bar', 'baz'])
[
{'confidence': 1, 'index': 1, 'value': 'bar'},
{'confidence': 0.6, 'index': 2, 'value': 'baz'}
]
``````

It does three things I really like:

1. It includes their indices in the result, so, I don’t have to worry about that fetching that manually.
2. It also includes results with similar characters, and gives them a confidence score.
3. Ranks the results according to their relevance.