问题
What is Big-O complexity of random.choice(list) in Python3, where n is amount of elements in a list ?
edit. Thank You all for give me the answer, now I understand.
回答1:
O(1). Or to be more precise, it's equivalent to the big-O random access time for looking up a single index in whatever sequence you pass it, and list has O(1) random access indexing (as does tuple). Simplified, all it does is seq[random.randrange(len(seq))], which is obviously equivalent to a single index lookup operation.
An example where it would be O(n) is collections.deque, where indexing in the middle of the deque is O(n) (with a largish constant divisor though, so it's not that expensive unless the deque is reaching the thousands of elements range or higher). So basically, don't use a deque if it's going to be large and you plan to select random elements from it repeatedly, stick to list, tuple, str, byte/bytearray, array.array and other sequence types with O(1) indexing.
回答2:
I think the above answer is incorrect. I empirically verified that the complexity of this operation is O(n). Here is my code and a little plot. I am not sure about the theory though.
from time import time
import numpy as np
import matplotlib.pyplot as plt
N = np.logspace(2, 10, 40)
output = []
for i, n in enumerate(N):
print(i)
n = int(n)
stats = time()
A = np.random.choice(list(range(n)), n//2)
output.append(time()-stats)
plt.plot(N, output)
This is the plot I got which looks quite linear to me.
回答3:
The complexity of random.choice(list) is O(log n) where n is the number of elements in the list.
The cpython implementation uses _randbelow(len(seq)) to get a pseudo-random index and then returns the item at that index.
The bottleneck is the _randbelow() function which uses rejection sampling to generate a number in the range [0, n). The function generates k pseudo-random bits with a call to getrandbits(k) where k is ceil(log N). These bits represent a number in the range [0, 2**k). This process is repeated until the generated number is less than n. Each call to the pseudo-random number generator runs in O(k) where k is the number of bits generated which is O(log n).
来源：https://stackoverflow.com/questions/40143157/q-what-is-big-o-complexity-of-random-choicelist-in-python3