Question

我有伪随机生成字符串的代码。如何让我的代码输出25个随机字符串？

import random, string

special=['@','!','#','$','%','^','&','*','(',')','-','=','+','{','[','}',']',';',':','.','>','<',',','?','/',]

def id_generator(size=25, chars=string.ascii_uppercase + string.digits + random.choice(special) ):
    return ''.join(random.choice(chars) for _ in range(size))

count = 0

def create_and_save_pseudorand():
    for count in range(100):
        return id_generator()
        #print(id_generator())

create_and_save_pseudorand()

Answer 1

我猜你要么

return [id_generator() for i in range(25)]

或将return与yield交换，然后使用：

for i in create_and_save_pseudorand():
    print i

btw：你似乎混淆了字符串的长度和数量。哪一个应该是25？

Answer 2

您的函数create_and_save_pseudorand当前在返回第一个随机生成的字符串后停止。将return更改为yield以解决此问题。然后，您可以引入count参数来确定要生成的随机字符串的数量。

def create_and_save_pseudorand(count):
    for count in range(count):
        yield id_generator()

然后，您可以使用

在控制台上打印结果

print(list(create_and_save_pseudorand(25)))

示例输出：

['YAB*MT7CU29IY12QMVLNH4QI9', '1E1445DCN*5K7RZIJNKSIKX10', '53W06LS8NJEAE0X60MFW*ES*W', 'KK8WMRA3DMPRWWYQDG2M0ED0Q', 'ZFLBQ0V04Z*E1PVUF4DG04IJ2', '292VPZA3QDKLQ4UBTKJA3TY3H', 'A77426HK1HNZVMUT883CZABIX', 'I2M4I0IO9GRLNVAH9BWV52AYJ', '*VUU3DGHN79SG5XPELHCKSG29', '*KWM2BUPGIL4OV3TUZ7614OAT', '29JZ*Y59*O2SPL7CWL9BB3*5V', 'IVZ0MK0HPIVCJG039Y9GTM*ZH', 'ZF2QSL5VXYW6B4*FOE4KRL4LW', 'ZOK*Z508DD1JE5UYHLGGKORKS', 'L2XIBU8J3IMZO9Y2GFLQSB3W7', 'PA6ZJGXS4G96FDA771143TOMM', '2ZHK13LVZZQ878L1OR2LQ3TGV', 'QGYR2WNJ58GPD0*54CHI6AKAR', 'TKPBD27HQ0TZFFZCVMIDAYJX4', 'G7HH9YSZ4P3ETRGY9REB90WDC', 'G3PTXR376AGJHX4V5V4NRLL30', 'XHM1J0A5Z48*58RFVH9W3Z5*O', 'XHG8VGNF*YA4GBX1SBE*DVGKV', '39YNN1X719N*WZ9VBUPLTVC5P', 'XYMJ57DH*3WX589J9ZENGFP0O']

生成25个伪随机字符串

2 个答案: