Python从int到string的快速转换

Question

我正在解决 python 中的大量阶乘，并发现当我计算完阶乘后，转换为字符串以保存到文件需要相同的时间。 我试图找到一种将 int 转换为字符串的快速方法。 我将举一个计算和int转换时间的例子。 我正在使用通用的 a = str(a) 但我觉得有更好的方法，比如使用缓冲区或库。

前任：

求解 100,000 阶乘 = 456,574 Digets

计算时间：6.36 秒

转换时间：5.20 秒

如果您有任何问题/解决方案，请告诉我！ 任何事情都会有所帮助。

import time

factorial = 1

print(" ")

one = int(input("lower  = "))
two = int(input("higher = "))

start = time.time()

for x in range(one,two + 1):
        factorial = factorial * two
        two = two - 1

end = time.time()

print("DONE! ")
print(end - start, "Seconds to compute")

start = time.time()

factorial = str(factorial)

f = open('Ans_1.txt','w')
f.write(factorial)
f.close()

end = time.time()

print(end - start, "Seconds to convert and save")

print(len(factorial), "Digets")

Answer 1

你可以尝试gmpy2 x.digits（[base]）。

import time
from gmpy2 import mpz


x = 123456789**12345

start = time.time()
python_str = str(x)
end = time.time()
print(end - start, "Python conversion time")

r = mpz(x)
start = time.time()
gmpy2_str = r.digits()
end = time.time()
print(end-start, "gmpy2 conversion time")

以上测试输出：

1.0336394309997559 Python转换时间

0.03306150436401367 gmpy2转换时间

Answer 2

这段代码更快（但还不够！：D）

结果：

╔═══╦════════════╦═════════════╦══════════════╦═══════════════════╗
║   ║ Count      ║ Compute(s)  ║  Convert(s)  ║  M.T Convert(s)   ║
╠═══╬════════════╬═════════════╬══════════════╬═══════════════════╣
║ 1 ║ 100,000    ║    2.68     ║     3.85     ║        2.81       ║
║ 2 ║ 250,000    ║   21.17     ║     39.83    ║       21.09       ║
╚═══╩════════════╩═════════════╩══════════════╩═══════════════════╝

无论如何，我认为你可以通过多线程更快地完成它。

导入时间导入数学导入线程

res_dict = {}

def int_str(threadID, each_thread, max_thread):
    if threadID == 1 :
        res_dict[threadID] = (str(factorial // 10 ** (each_thread * (max_thread - 1))))
    elif threadID == max_thread:
        res_dict[threadID] = (str(int(factorial % 10 ** (each_thread * 1))))
    else: 
        tmp = (factorial % 10 ** (each_thread * (max_thread - threadID + 1))) // 10 ** (each_thread * (max_thread - threadID))
        pre = "0" * ((digits // max_thread) - (math.floor(math.log10(tmp))+1))
        res_dict[threadID] = (pre + str(int(tmp)))

factorial = 1

print(" ")

def fact(a,b):
    if b == 1:
        return 1
    else:
        return a * fact(a,b-1)

one = int(input("lower  = "))
two = int(input("higher = "))

start = time.time()

for x in range(one,two + 1):
        factorial = factorial * two
        two = two - 1

end = time.time()

print("DONE! ")
print(end - start, "Seconds to compute")


start = time.time()

digits = math.floor(math.log10(factorial))+1

max_thread      = 3
each_thread     = digits // max_thread

tr = []

for item in range(1, max_thread + 1):
    t = threading.Thread(target=int_str, args=(item, each_thread, max_thread))
    t.start()
    tr.append(t)


for item in tr:
    item.join()

last_res = ''

for item in sorted(res_dict):
    if item != max_thread:
        last_res += res_dict[item]
    else:
        last_res += ("0" * (digits - len(last_res) - len(res_dict[item]))) + res_dict[item]


f = open('Ans_2.txt','w')
f.write(last_res)
f.close()


end = time.time()
print(end - start, "Seconds to convert and save")

print(digits, "Digets")

更新：

只需用pypy运行你的代码就会非常快！

╔═══╦════════════╦═════════════╦══════════════╦═══════════════════╗
║   ║ Count      ║ Compute(s)  ║  Convert(s)  ║ pypy Convert(s)   ║
╠═══╬════════════╬═════════════╬══════════════╬═══════════════════╣
║ 1 ║ 100,000    ║    2.98     ║     3.85     ║        0.79       ║
║ 2 ║ 250,000    ║   25.83     ║     39.83    ║        7.17       ║
╚═══╩════════════╩═════════════╩══════════════╩═══════════════════╝

Answer 3

int().__str__()比str(int)快，因为str()需要找到它正在转换的类型。

import time
    
start=time.time()
x1=[x.__str__() for x in range(10000)]
print(f"__str__: {time.time()-start}")

start=time.time()
x2=[str(x) for x in range(10000)]
print(f"str: {time.time()-start}")

print(x1==x2)

>>>__str__: 0.005000591278076172
>>>str: 0.006000518798828125
>>>True