def gen_random_tufo(form): iden = hexlify(random.bytes(16)).decode('utf8') props = {} for propname in form: if random.random() <= INTEGER_VAL_RATE: val = random.randint(-2 ** 62, 2 ** 63) else: val = random_string(random_val_len()) props[propname] = val return (iden, props)
def test_base64(): random.seed(73101) for i in xrange(6): for n in xrange(20): s = random.bytes(n) ours = base64_encode(s) theirs = base64.b64encode(s) assert ours==theirs assert base64_decode(ours)==s for s in '====','&&&&','++==','+++=': try: base64_decode(s) except ValueError: pass
def get_random_bytes(size, seed): random.seed(seed) return random.bytes(size)
def test_bytes(self): random.seed(101) self.assertEqual(random.bytes(10), "_\xb32\x84\x0bFs\x8dQE")
from numpy.random import bytes print repr(bytes(5)) # string of 5 random bytes print repr(bytes(5)) # another string of 5 random bytes
def rand_chars(): n_bytes = random.randint(0, 10) return random.bytes(n_bytes).decode("latin1")
'''[[-0.84341992 -0.51332541] [-0.0435368 -0.27530091] [-1.56306679 -0.79563099] [ 0.9305844 0.67776741] [ 0.69844026 0.17360206]]''' x = random.normal(size=(5, 2)) # 产生5个,n=5,p=0.5的二项分布样本 x = random.binomial(n=5, p=0.5, size=5) print(x) a = np.arange(10) print(a) # 从a中有回放的随机采样7个 x = random.choice(a, 7) print(x) # 从a中无回放的随机采样7个 x = random.choice(a, 7, replace=False) print(x) # 对a进行乱序并返回一个新的array b = random.permutation(a) print(b) # 对a进行in-place乱序 print('start', a) random.shuffle(a) print('end', a) # 生成一个长度为9的随机bytes序列并作为str返回 # '\x96\x9d\xd1?\xe6\x18\xbb\x9a\xec' x = random.bytes(9) print(x)
import sys sys.path.append('../cryptoutils/') import AES import binascii import numpy.random as rand import time secret = binascii.a2b_base64( 'Um9sbGluJyBpbiBteSA1LjAKV2l0aCBteSByYWctdG9wIGRvd24gc28gbXkgUm9sbGluJyBpbiBteSA1LjAKV2l0aCBteSByYWctdG9wIGRvd24gc28gbXkgaGFpciBjYW4gYmxvdwpUaGUgZ2lybGllcyBvbiBzdGFuZGJ5IHdhdmluZyBqdXN0IHRvIHNheSBoaQpEaWQgeW91IHN0b3A/IE5vLCBJIGp1c3QgZHJvdmUgYnkK' ) key = rand.bytes(16) prefix_len = rand.randint(1, 16) prefix = rand.bytes(prefix_len) def oracle(s): plain = prefix + s + secret return AES.ecb_encrypt(plain, key) """ This is not a clean solution but it is a soultion i am proud of it doesn't break any assumptions (i guess i don't find block size but hey) it is a huge advancement over break_ecb """ def break_ecb():
def get_random_bytes(size, seed): with local_seed(seed): return random.bytes(size)