检查列表中是否存在值的最快方法

Question

What is the fastest way to check if a value exists in a very large list?

Answer 1

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)

Answer 2

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 too long or up if there are too 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()

Answer 3

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.

Answer 4

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.

Answer 5

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()

Answer 6

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

index = dict((y,x) for x,y in enumerate(a))
try:
   a_index = index[7]
except KeyError:
   print "Not found"
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.

Answer 7

Be aware that the in operator tests not only equality ( == ) but also identity ( is ), the in logic for list s 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 list s 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 ).

Answer 8

If you only want to check the existence of one element in a list,

7 in list_data

is the fastest solution. Note though that

7 in set_data

is a near-free operation, independently of the size of the set! Creating a set from a large list is 300 to 400 times slower than in , so if you need to check for many elements, creating a set first is faster.

Plot created with perfplot :

import perfplot
import numpy as np


def setup(n):
    data = np.arange(n)
    np.random.shuffle(data)
    return data, set(data)


def list_in(data):
    return 7 in data[0]


def create_set_from_list(data):
    return set(data[0])


def set_in(data):
    return 7 in data[1]


b = perfplot.bench(
    setup=setup,
    kernels=[list_in, set_in, create_set_from_list],
    n_range=[2 ** k for k in range(24)],
    xlabel="len(data)",
    equality_check=None,
)
b.save("out.png")
b.show()

Answer 9

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.

Answer 10

Or use __contains__ :

sequence.__contains__(value)

Demo:

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

Answer 11

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.

Answer 12

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.

Answer 13

@Winston Ewert's solution yields a big speed-up for very large lists, but this stackoverflow answer indicates that the the try:/except:/else: construct will be slowed down if the except branch is often reached. An alternative is to take advantage of the .get() method for the dict:

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

index = dict((y, x) for x, y in enumerate(a))

b = index.get(7, None)
if b is not None:
    "Do something with variable b"

The .get(key, default) method is just for the case when you can't guarantee a key will be in the dict. If key is present, it returns the value (as would dict[key] ), but when it is not, .get() returns your default value (here None ). You need to make sure in this case that the chosen default will not be in a .

Answer 14

present = False
searchItem = 'd'
myList = ['a', 'b', 'c', 'd', 'e']
if searchItem in myList:
   present = True
   print('present = ', present)
else:
   print('present = ', present)

Answer 15

i think it's good

mylist = [j for j in range(100)]
value = 13 #mutable
print (value in mylist)
#output: True

if you wanna print the value:

mylist = [j for j in range(100)]
value = 13 #mutable
if value in mylist:
    print (value)

Answer 16

Edge case for spatial data

There are probably faster algorithms for handling spatial data (eg refactoring to use a kd tree), but the special case of checking if a vector is in an array is useful:

• If you have spatial data (ie cartesian coordinates)
• If you have integer masks (ie array filtering)

In this case, I was interested in knowing if an (undirected) edge defined by two points was in a collection of (undirected) edges, such that

(pair in unique_pairs) | (pair[::-1] in unique_pairs) for pair in pairs

where pair constitutes two vectors of arbitrary length (ie shape (2,N) ).

If the distance between these vectors is meaningful, then the test can be expressed by a floating point inequality like

test_result = Norm(v1 - v2) < Tol

and "Value exists in List" is simply any(test_result) .

Example code and dummy test set generators for integer pairs and R3 vector pairs are below.

# 3rd party
import numpy as np
import numpy.linalg as LA
import matplotlib.pyplot as plt

# optional
try:
    from tqdm import tqdm
except ModuleNotFoundError:
    def tqdm(X, *args, **kwargs):
        return X
    print('tqdm not found. tqdm is a handy progress bar module.')
    

def get_float_r3_pairs(size):
    """ generate dummy vector pairs in R3  (i.e. case of spatial data) """
    coordinates = np.random.random(size=(size, 3))
    pairs = []
    for b in coordinates:
        for a in coordinates:
            pairs.append((a,b))
    pairs = np.asarray(pairs)
    return pairs
    
        
def get_int_pairs(size):
    """ generate dummy integer pairs (i.e. case of array masking) """
    coordinates = np.random.randint(0, size, size)
    pairs = []
    for b in coordinates:
        for a in coordinates:
            pairs.append((a,b))
    pairs = np.asarray(pairs)
    return pairs


def float_tol_pair_in_pairs(pair:np.ndarray, pairs:np.ndarray) -> np.ndarray:
    """
    True if abs(a0 - b0) <= tol & abs(a1 - b1) <= tol for (ai1, aj2), (bi1, bj2)
    in [(a01, a02), ... (aik, ajl)]
    
    NB this is expected to be called in iteration so no sanitization is performed.

    Parameters
    ----------
    pair : np.ndarray
        pair of vectors with shape (2, M)
    pairs : np.ndarray
        collection of vector pairs with shape (N, 2, M)

    Returns
    -------
    np.ndarray
        (pair in pairs) | (pair[::-1] in pairs).
    """
    m1 = np.sum( abs(LA.norm(pairs - pair, axis=2)) <= (1e-03, 1e-03), axis=1 ) == 2
    m2 = np.sum( abs(LA.norm(pairs - pair[::-1], axis=2)) <= (1e-03, 1e-03), axis=1 ) == 2
    return m1 | m2


def get_unique_pairs(pairs:np.ndarray) -> np.ndarray:
    """
    apply float_tol_pair_in_pairs for pair in pairs
    
    Parameters
    ----------
    pairs : np.ndarray
        collection of vector pairs with shape (N, 2, M)

    Returns
    -------
    np.ndarray
        pair if not ((pair in rv) | (pair[::-1] in rv)) for pair in pairs

    """
    pairs = np.asarray(pairs).reshape((len(pairs), 2, -1))
    rv = [pairs[0]]
    for pair in tqdm(pairs[1:], desc='finding unique pairs...'):
        if not any(float_tol_pair_in_pairs(pair, rv)):
            rv.append(pair)
    return np.array(rv)

Answer 17

For me it was 0.030 sec (real), 0.026 sec (user), and 0.004 sec (sys).

try:
print("Started")
x = ["a", "b", "c", "d", "e", "f"]

i = 0

while i < len(x):
    i += 1
    if x[i] == "e":
        print("Found")
except IndexError:
    pass

Answer 18

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])