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[英]How to Generate random code with 4 digit string and 4 digit number in Python?
[英]Python: How to generate a 12-digit random number?
在Python中,如何生成一個12位的隨機數? 是否有任何函數可以指定像random.range(12)
這樣的范圍?
import random
random.randint()
輸出應該是一個包含 0-9 范圍內的 12 位數字的字符串(允許前導零)。
直接的方法有什么問題?
>>> import random
>>> random.randint(100000000000,999999999999)
544234865004L
如果您希望它帶有前導零,則需要一個字符串。
>>> "%0.12d" % random.randint(0,999999999999)
'023432326286'
編輯:
我自己對這個問題的解決方案是這樣的:
import random
def rand_x_digit_num(x, leading_zeroes=True):
"""Return an X digit number, leading_zeroes returns a string, otherwise int"""
if not leading_zeroes:
# wrap with str() for uniform results
return random.randint(10**(x-1), 10**x-1)
else:
if x > 6000:
return ''.join([str(random.randint(0, 9)) for i in xrange(x)])
else:
return '{0:0{x}d}'.format(random.randint(0, 10**x-1), x=x)
測試結果:
>>> rand_x_digit_num(5)
'97225'
>>> rand_x_digit_num(5, False)
15470
>>> rand_x_digit_num(10)
'8273890244'
>>> rand_x_digit_num(10)
'0019234207'
>>> rand_x_digit_num(10, False)
9140630927L
速度計時方法:
def timer(x):
s1 = datetime.now()
a = ''.join([str(random.randint(0, 9)) for i in xrange(x)])
e1 = datetime.now()
s2 = datetime.now()
b = str("%0." + str(x) + "d") % random.randint(0, 10**x-1)
e2 = datetime.now()
print "a took %s, b took %s" % (e1-s1, e2-s2)
速度測試結果:
>>> timer(1000)
a took 0:00:00.002000, b took 0:00:00
>>> timer(10000)
a took 0:00:00.021000, b took 0:00:00.064000
>>> timer(100000)
a took 0:00:00.409000, b took 0:00:04.643000
>>> timer(6000)
a took 0:00:00.013000, b took 0:00:00.012000
>>> timer(2000)
a took 0:00:00.004000, b took 0:00:00.001000
它告訴我們什么:
對於長度低於 6000 個字符的任何數字,我的方法更快——有時快得多,但對於更大的數字, arshajii建議的方法看起來更好。
做random.randrange(10**11, 10**12)
。 它的工作原理就像randint
滿足range
從文檔:
randrange(self, start, stop=None, step=1, int=<type 'int'>, default=None, maxwidth=9007199254740992L) method of random.Random instance
Choose a random item from range(start, stop[, step]).
This fixes the problem with randint() which includes the
endpoint; in Python this is usually not what you want.
Do not supply the 'int', 'default', and 'maxwidth' arguments.
這實際上就像做random.choice(range(10**11, 10**12))
或random.randint(10**1, 10**12-1)
。 由於它符合與range()
相同的語法,因此它比這兩個替代方案更直觀和更清晰
如果允許前導零:
"%012d" %random.randrange(10**12)
由於允許前導零(根據您的評論),您還可以使用:
int(''.join(str(random.randint(0,9)) for _ in xrange(12)))
編輯:當然,如果你想要一個string ,你可以省略int
部分:
''.join(str(random.randint(0,9)) for _ in xrange(12))
在我看來,這似乎是最直接的方法。
有很多方法可以做到這一點:
import random
rnumber1 = random.randint(10**11, 10**12-1) # randint includes endpoint
rnumber2 = random.randrange(10**11, 10**12) # randrange does not
# useful if you want to generate some random string from your choice of characters
digits = "123456789"
digits_with_zero = digits + "0"
rnumber3 = random.choice(digits) + ''.join(random.choice(digits_with_zero) for _ in range(11))
from random import randint
def random_with_N_digits(n):
range_start = 10**(n-1)
range_end = (10**n)-1
return randint(range_start, range_end)
print random_with_N_digits(12)
import random
def generate_random_digits(n):
if not n > 0:
return None
randnum = random.randint(10**n, 2*10**n)
return str(randnum)[-n:]
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