python – Fastest way to check if a value exists in a list

The Question :

908 people think this question is useful

What is the fastest way to know if a value exists in a list (a list with millions of values in it) and what its index is?

I know that all values in the list are unique as in this example.

The first method I try is (3.8 sec in my real code):

a = [4,2,3,1,5,6]

if a.count(7) == 1:
b=a.index(7)
"Do something with variable b"



The second method I try is (2x faster: 1.9 sec for my real code):

a = [4,2,3,1,5,6]

try:
b=a.index(7)
except ValueError:
"Do nothing"
else:
"Do something with variable b"



Proposed methods from Stack Overflow user (2.74 sec for my real code):

a = [4,2,3,1,5,6]
if 7 in a:
a.index(7)



In my real code, the first method takes 3.81 sec and the second method takes 1.88 sec. It’s a good improvement, but:

I’m a beginner with Python/scripting, and is there a faster way to do the same things and save more processing time?

More specific explanation for my application:

In the Blender API I can access a list of particles:

particles = [1, 2, 3, 4, etc.]



From there, I can access a particle’s location:

particles[x].location = [x,y,z]



And for each particle I test if a neighbour exists by searching each particle location like so:

if [x+1,y,z] in particles.location
"Find the identity of this neighbour particle in x:the particle's index
in the array"
particles.index([x+1,y,z])


• In python the thing in square brackets is called a list, not an array. Rather than using a list use a set. Or keep your list sorted and use the bisect module
• So you really need to juggle indices? Or doesn’t order actually matter and you just want to do member ship tests, intersections, etc.? In order words, it depends on what you’re really trying to do. Sets may work for you, and then they are a really good answer, but we can’t tell from the code you showed.
• Probably you have to specify in your question that you need not the value, but its index.
• I edit my question and try to explain more clearly what I want to do … I hope so…
• @StevenRumbalski: because set cannot contain duplication content, while Jean wants to store location of particles (x,y,z could be the same), we cannot use set in this case

1721 people think this answer is useful
7 in a



Clearest and fastest way to do it.

You can also consider using a set, but constructing that set from your list may take more time than faster membership testing will save. The only way to be certain is to benchmark well. (this also depends on what operations you require)

243 people think this answer is useful

As stated by others, in can be very slow for large lists. Here are some comparisons of the performances for in, set and bisect. Note the time (in second) is in log scale.

Code for testing:

import random
import bisect
import matplotlib.pyplot as plt
import math
import time

def method_in(a, b, c):
start_time = time.time()
for i, x in enumerate(a):
if x in b:
c[i] = 1
return time.time() - start_time

def method_set_in(a, b, c):
start_time = time.time()
s = set(b)
for i, x in enumerate(a):
if x in s:
c[i] = 1
return time.time() - start_time

def method_bisect(a, b, c):
start_time = time.time()
b.sort()
for i, x in enumerate(a):
index = bisect.bisect_left(b, x)
if index < len(a):
if x == b[index]:
c[i] = 1
return time.time() - start_time

def profile():
time_method_in = []
time_method_set_in = []
time_method_bisect = []

# adjust range down if runtime is to great or up if there are to many zero entries in any of the time_method lists
Nls = [x for x in range(10000, 30000, 1000)]
for N in Nls:
a = [x for x in range(0, N)]
random.shuffle(a)
b = [x for x in range(0, N)]
random.shuffle(b)
c = [0 for x in range(0, N)]

time_method_in.append(method_in(a, b, c))
time_method_set_in.append(method_set_in(a, b, c))
time_method_bisect.append(method_bisect(a, b, c))

plt.plot(Nls, time_method_in, marker='o', color='r', linestyle='-', label='in')
plt.plot(Nls, time_method_set_in, marker='o', color='b', linestyle='-', label='set')
plt.plot(Nls, time_method_bisect, marker='o', color='g', linestyle='-', label='bisect')
plt.xlabel('list size', fontsize=18)
plt.ylabel('log(time)', fontsize=18)
plt.legend(loc='upper left')
plt.yscale('log')
plt.show()

profile()



42 people think this answer is useful

You could put your items into a set. Set lookups are very efficient.

Try:

s = set(a)
if 7 in s:
# do stuff



edit In a comment you say that you’d like to get the index of the element. Unfortunately, sets have no notion of element position. An alternative is to pre-sort your list and then use binary search every time you need to find an element.

31 people think this answer is useful
def check_availability(element, collection: iter):
return element in collection



Usage

check_availability('a', [1,2,3,4,'a','b','c'])



I believe this is the fastest way to know if a chosen value is in an array.

20 people think this answer is useful

The original question was:

What is the fastest way to know if a value exists in a list (a list with millions of values in it) and what its index is?

Thus there are two things to find:

1. is an item in the list, and
2. what is the index (if in the list).

Towards this, I modified @xslittlegrass code to compute indexes in all cases, and added an additional method.

Results

Methods are:

1. in–basically if x in b: return b.index(x)
2. try–try/catch on b.index(x) (skips having to check if x in b)
3. set–basically if x in set(b): return b.index(x)
4. bisect–sort b with its index, binary search for x in sorted(b). Note mod from @xslittlegrass who returns the index in the sorted b, rather than the original b)
5. reverse–form a reverse lookup dictionary d for b; then d[x] provides the index of x.

Results show that method 5 is the fastest.

Interestingly the try and the set methods are equivalent in time.

Test Code

import random
import bisect
import matplotlib.pyplot as plt
import math
import timeit
import itertools

def wrapper(func, *args, **kwargs):
" Use to produced 0 argument function for call it"
# Reference https://www.pythoncentral.io/time-a-python-function/
def wrapped():
return func(*args, **kwargs)
return wrapped

def method_in(a,b,c):
for i,x in enumerate(a):
if x in b:
c[i] = b.index(x)
else:
c[i] = -1
return c

def method_try(a,b,c):
for i, x in enumerate(a):
try:
c[i] = b.index(x)
except ValueError:
c[i] = -1

def method_set_in(a,b,c):
s = set(b)
for i,x in enumerate(a):
if x in s:
c[i] = b.index(x)
else:
c[i] = -1
return c

def method_bisect(a,b,c):
" Finds indexes using bisection "

# Create a sorted b with its index
bsorted = sorted([(x, i) for i, x in enumerate(b)], key = lambda t: t[0])

for i,x in enumerate(a):
index = bisect.bisect_left(bsorted,(x, ))
c[i] = -1
if index < len(a):
if x == bsorted[index][0]:
c[i] = bsorted[index][1]  # index in the b array

return c

def method_reverse_lookup(a, b, c):
reverse_lookup = {x:i for i, x in enumerate(b)}
for i, x in enumerate(a):
c[i] = reverse_lookup.get(x, -1)
return c

def profile():
Nls = [x for x in range(1000,20000,1000)]
number_iterations = 10
methods = [method_in, method_try, method_set_in, method_bisect, method_reverse_lookup]
time_methods = [[] for _ in range(len(methods))]

for N in Nls:
a = [x for x in range(0,N)]
random.shuffle(a)
b = [x for x in range(0,N)]
random.shuffle(b)
c = [0 for x in range(0,N)]

for i, func in enumerate(methods):
wrapped = wrapper(func, a, b, c)
time_methods[i].append(math.log(timeit.timeit(wrapped, number=number_iterations)))

markers = itertools.cycle(('o', '+', '.', '>', '2'))
colors = itertools.cycle(('r', 'b', 'g', 'y', 'c'))
labels = itertools.cycle(('in', 'try', 'set', 'bisect', 'reverse'))

for i in range(len(time_methods)):
plt.plot(Nls,time_methods[i],marker = next(markers),color=next(colors),linestyle='-',label=next(labels))

plt.xlabel('list size', fontsize=18)
plt.ylabel('log(time)', fontsize=18)
plt.legend(loc = 'upper left')
plt.show()

profile()



19 people think this answer is useful
a = [4,2,3,1,5,6]

index = dict((y,x) for x,y in enumerate(a))
try:
a_index = index[7]
except KeyError:
else:
print "found"



This will only be a good idea if a doesn’t change and thus we can do the dict() part once and then use it repeatedly. If a does change, please provide more detail on what you are doing.

8 people think this answer is useful

It sounds like your application might gain advantage from the use of a Bloom Filter data structure.

In short, a bloom filter look-up can tell you very quickly if a value is DEFINITELY NOT present in a set. Otherwise, you can do a slower look-up to get the index of a value that POSSIBLY MIGHT BE in the list. So if your application tends to get the “not found” result much more often then the “found” result, you might see a speed up by adding a Bloom Filter.

For details, Wikipedia provides a good overview of how Bloom Filters work, and a web search for “python bloom filter library” will provide at least a couple useful implementations.

7 people think this answer is useful

Be aware that the in operator tests not only equality (==) but also identity (is), the in logic for lists is roughly equivalent to the following (it’s actually written in C and not Python though, at least in CPython):

for element in s:
if element is target:
# fast check for identity implies equality
return True
if element == target:
# slower check for actual equality
return True
return False



In most circumstances this detail is irrelevant, but in some circumstances it might leave a Python novice surprised, for example, numpy.NAN has the unusual property of being not being equal to itself:

>>> import numpy
>>> numpy.NAN == numpy.NAN
False
>>> numpy.NAN is numpy.NAN
True
>>> numpy.NAN in [numpy.NAN]
True



To distinguish between these unusual cases you could use any() like:

>>> lst = [numpy.NAN, 1 , 2]
>>> any(element == numpy.NAN for element in lst)
False
>>> any(element is numpy.NAN for element in lst)
True



Note the in logic for lists with any() would be:

any(element is target or element == target for element in lst)



However, I should emphasize that this is an edge case, and for the vast majority of cases the in operator is highly optimised and exactly what you want of course (either with a list or with a set).

2 people think this answer is useful

Or use __contains__:

sequence.__contains__(value)



Demo:

>>> l=[1,2,3]
>>> l.__contains__(3)
True
>>>



2 people think this answer is useful

This is not the code, but the algorithm for very fast searching.

If your list and the value you are looking for are all numbers, this is pretty straightforward. If strings: look at the bottom:

• -Let “n” be the length of your list
• -Optional step: if you need the index of the element: add a second column to the list with current index of elements (0 to n-1) – see later
• Order your list or a copy of it (.sort())
• Loop through:
• Compare your number to the n/2th element of the list
• If larger, loop again between indexes n/2-n
• If smaller, loop again between indexes 0-n/2
• If the same: you found it
• Keep narrowing the list until you have found it or only have 2 numbers (below and above the one you are looking for)
• This will find any element in at most 19 steps for a list of 1.000.000 (log(2)n to be precise)

If you also need the original position of your number, look for it in the second, index column.

If your list is not made of numbers, the method still works and will be fastest, but you may need to define a function which can compare/order strings.

Of course, this needs the investment of the sorted() method, but if you keep reusing the same list for checking, it may be worth it.

1 people think this answer is useful

Because the question is not always supposed to be understood as the fastest technical way – I always suggest the most straightforward fastest way to understand/write: a list comprehension, one-liner

[i for i in list_from_which_to_search if i in list_to_search_in]



I had a list_to_search_in with all the items, and wanted to return the indexes of the items in the list_from_which_to_search.

This returns the indexes in a nice list.

There are other ways to check this problem – however list comprehensions are quick enough, adding to the fact of writing it quick enough, to solve a problem.

-2 people think this answer is useful

Code to check whether two elements exist in array whose product equals k:

n = len(arr1)
for i in arr1:
if k%i==0:
print(i)