Comment by Jtsummers

I'm not a numpy expert, I just use it on occasion but this works and eliminates the explicit inner loop, but not the outer loop. It collects the list of prime numbers while removing multiples of the prime from the numpy array of numbers:

import numpy as np

  def sieve_primes(limit):
    nums = np.arange(limit + 1)
    nums[1] = 0
    primes = []
    for i in range(2, limit):
      if nums[i] != 0: # could just be `if nums[i]:` since 0 is false-y
        primes.append(i)
        nums = nums * (np.mod(nums, i) != 0)
    print(primes)
  sieve_primes(1000)

This takes advantage of the fact that True and False are treated as 1 and 0 in Python for the multiplication.

EDIT: The other solutions people have shown are closer to the sieve itself than my solution. I was having trouble with the slicing notation, now that I've seen how that should be done I'd redo the above as:

  def sieve_primes(limit):
    nums = np.arange(limit + 1)
    nums[1] = 0
    for i in range(2, limit):
      if nums[i] != 0:
        nums[i+i::i] = 0
    print(np.nonzero(nums))

However, the first version is faster by my testing so I'd end up just going back to it if I really cared about performance.