在python中随机播放字符串数据

Question

我有一列有1000万个字符串。字符串中的字符需要以某种方式重新排列。

原始字串： AAA01188P001

乱序字符串： 188A1A0AP001

现在，我正在运行一个for循环，该循环将接收每个字符串并重新定位每个字母，但这需要几个小时才能完成。有没有更快的方法来达到这个结果？

这是for循环。

for i in range(0, len(OrderProduct)):
    s = list(OrderProduct['OrderProductId'][i])
    a = s[1]
    s[1] = s[7]
    s[7] = a 
    a = s[3]
    s[3] = s[6]
    s[6] = a 
    a = s[2]
    s[2] = s[3]
    s[3] = a 
    a = s[5]
    s[5] = s[0]
    s[0] = a 
    OrderProduct['OrderProductId'][i] = ''.join(s)

Answer 1

我使用不同的方法进行了一些性能测试：

这是我获得1000000次随机播放的结果：

188A1AA0P001 usefString 0.518183742
188A1AA0P001 useMap     1.415851829
188A1AA0P001 useConcat  0.5654986979999999
188A1AA0P001 useFormat  0.800639699
188A1AA0P001 useJoin    0.5488918539999998

基于此，带有硬编码子字符串的格式字符串似乎是最快的。

这是我用来测试的代码：

def usefString(s): return f"{s[5:8]}{s[0]}{s[4]}{s[1:4]}{s[8:]}"

posMap = [5,6,7,0,4,1,2,3,8,9,10,11]
def useMap(s): return "".join(map(lambda i:s[i], posMap))

def useConcat(s): return s[5:8]+s[0]+s[4]+s[1:4]+s[8:]

def useFormat(s): return '{}{}{}{}{}'.format(s[5:8],s[0],s[4],s[1:4],s[8:])

def useJoin(s): return "".join([s[5:8],s[0],s[4],s[1:4],s[8:]])

from timeit import timeit
count = 1000000
s = "AAA01188P001"

t = timeit(lambda:usefString(s),number=count)
print(usefString(s),"usefString",t)

t = timeit(lambda:useMap(s),number=count)
print(useMap(s),"useMap",t)

t = timeit(lambda:useConcat(s),number=count)
print(useConcat(s),"useConcat",t)

t = timeit(lambda:useFormat(s),number=count)
print(useFormat(s),"useFormat",t)

t = timeit(lambda:useJoin(s),number=count)
print(useJoin(s),"useJoin",t)

表演：（由@jezrael添加）

N = 1000000
OrderProduct = pd.DataFrame({'OrderProductId':['AAA01188P001'] * N})

In [331]: %timeit [f'{s[5:8]}{s[0]}{s[4]}{s[1:4]}{s[8:]}' for s in OrderProduct['OrderProductId']]
527 ms ± 16.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [332]: %timeit [s[5:8]+s[0]+s[4]+s[1:4]+s[8:] for s in OrderProduct['OrderProductId']]
610 ms ± 18.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [333]: %timeit ['{}{}{}{}{}'.format(s[5:8],s[0],s[4],s[1:4],s[8:]) for s in OrderProduct['OrderProductId']]
954 ms ± 76.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [334]: %timeit ["".join([s[5:8],s[0],s[4],s[1:4],s[8:]]) for s in OrderProduct['OrderProductId']]
594 ms ± 10.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)