# Fastest way to check if a value exists in a list

`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)

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 randomimport bisectimport matplotlib.pyplot as pltimport mathimport timedef 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_timedef 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_timedef 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_timedef 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()`

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.