在Python中,如何找到整数中的位数?
当前回答
这里是最简单的方法,不需要将int转换为字符串:
假设给出的数字为15位,例如;n = 787878899999999;
n=787878899999999
n=abs(n) // we are finding absolute value because if the number is negative int to string conversion will produce wrong output
count=0 //we have taken a counter variable which will increment itself till the last digit
while(n):
n=n//10 /*Here we are removing the last digit of a number...it will remove until 0 digits will left...and we know that while(0) is False*/
count+=1 /*this counter variable simply increase its value by 1 after deleting a digit from the original number
print(count) /*when the while loop will become False because n=0, we will simply print the value of counter variable
输入:
n=787878899999999
输出:
15
其他回答
from math import log10
digits = lambda n: ((n==0) and 1) or int(log10(abs(n)))+1
顶部的答案是说mathlog10更快,但我得到的结果表明len(str(n))更快。
arr = []
for i in range(5000000):
arr.append(random.randint(0,12345678901234567890))
%%timeit
for n in arr:
len(str(n))
//2.72 s ± 304 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%%timeit
for n in arr:
int(math.log10(n))+1
//3.13 s ± 545 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
此外,我没有在数学方法中添加逻辑来返回准确的结果,我只能想象这会使它更加缓慢。
我不知道之前的答案是如何证明数学方法更快的。
不需要转换为字符串
import math
digits = int(math.log10(n))+1
也可以处理0和负数
import math
if n > 0:
digits = int(math.log10(n))+1
elif n == 0:
digits = 1
else:
digits = int(math.log10(-n))+2 # +1 if you don't count the '-'
你可能想把它放在一个函数中:)
以下是一些基准测试。len(str())对于非常小的数字已经落后了
timeit math.log10(2**8)
1000000 loops, best of 3: 746 ns per loop
timeit len(str(2**8))
1000000 loops, best of 3: 1.1 µs per loop
timeit math.log10(2**100)
1000000 loops, best of 3: 775 ns per loop
timeit len(str(2**100))
100000 loops, best of 3: 3.2 µs per loop
timeit math.log10(2**10000)
1000000 loops, best of 3: 844 ns per loop
timeit len(str(2**10000))
100 loops, best of 3: 10.3 ms per loop
好吧,如果不转换为字符串,我会这样做:
def lenDigits(x):
"""
Assumes int(x)
"""
x = abs(x)
if x < 10:
return 1
return 1 + lenDigits(x / 10)
最小递归FTW
这里是最简单的方法,不需要将int转换为字符串:
假设给出的数字为15位,例如;n = 787878899999999;
n=787878899999999
n=abs(n) // we are finding absolute value because if the number is negative int to string conversion will produce wrong output
count=0 //we have taken a counter variable which will increment itself till the last digit
while(n):
n=n//10 /*Here we are removing the last digit of a number...it will remove until 0 digits will left...and we know that while(0) is False*/
count+=1 /*this counter variable simply increase its value by 1 after deleting a digit from the original number
print(count) /*when the while loop will become False because n=0, we will simply print the value of counter variable
输入:
n=787878899999999
输出:
15
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