How can I select items smaller than a given integer from a sorted list?

Question

I have an array of primes, for example. between integers from 0 to 1000

primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

I get input

n = int(input())

What is the most efficient way to slice an array into a new array, where the last element of the array will be smaller n?

Answer 1

You can use the fact that it is primesalready sorted bisect, like this

>>> from bisect import bisect
>>> primes[:bisect(primes, n)]

bisectperforms a binary search in the input list and returns the index of an element that is smaller n.

Answer 2

list, . , , , . , , , :

small_primes = primes[:bisect.bisect(primes, n)]

NumPy, , , , , . , primes ndarray, , , O (log N).

small_primes = primes[:bisect.bisect(primes, n)]

"" , , :

small_primes = itertools.takewhile(lambda p: p<n, primes)

0; , , . , , - .

, " " , , , NumPy PyPy, ...

Answer 3

, " " (? ? - ?), , , , :

168 . , . , N = 168, , zillions , 1000 . , :

prime_slices = [[prime for prime in primes if prime < n] for n in range(1000)]

, n:

prime_slices[n]

. , O (N ^ 2), , , O (N) . O (N ^ 2) , 15K--, .

Answer 4

- , :

primes[bisect(primes, n):]=[]

, , , , .