如何获取数字的前两位？

Question

我想检查 Python 中数字的前两位。像这样的：

for i in range(1000):

    if(first two digits of i == 15):
        print("15")

    elif(first two digits of i == 16):
        print("16")

是否有检查数字前两位数字的命令？我想避免像if(i>149 and i<160):...这样的命令

Answer 1

将 O(n) 时间解决方案与其他答案中提供的“恒定时间”O(1) 解决方案进行比较表明，如果 O(n) 算法足够快，n 可能在它比慢 O(1) 慢之前必须变得非常大。

strings 版本约为。对于 20 位或更少位数的数字，比“数学”版本快 60%。只有当位数接近 200 位数

时，它们才会变得更接近

# the "maths" version
import math

def first_n_digits1(num, n):
    return num // 10 ** (int(math.log(num, 10)) - n + 1)

%timeit first_n_digits1(34523452452, 2)
1.21 µs ± 75 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit first_n_digits1(34523452452, 8)
1.24 µs ± 47.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

# 22 digits
%timeit first_n_digits1(3423234239472523452452, 2)
1.33 µs ± 59.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit first_n_digits1(3423234239472523452452, 15)
1.23 µs ± 61.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

# 196 digits
%timeit first_n_digits1(3423234239472523409283475908723908723409872390871243908172340987123409871234012089172340987734507612340981344509873123401234670350981234098123140987314509812734091823509871345109871234098172340987125988123452452, 39)
1.86 µs ± 21.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

# The "string" verions
def first_n_digits2(num, n):
    return int(str(num)[:n])

%timeit first_n_digits2(34523452452, 2)
744 ns ± 28.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit first_n_digits2(34523452452, 8)
768 ns ± 42.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

# 22 digits
%timeit first_n_digits2(3423234239472523452452, 2)
767 ns ± 33.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit first_n_digits2(3423234239472523452452, 15)
830 ns ± 55.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

# 196 digits
%timeit first_n_digits2(3423234239472523409283475908723908723409872390871243908098712340987123401208917234098773450761234098134450987312340123467035098123409812314098734091823509871345109871234098172340987125988123452452, 39)
1.87 µs ± 140 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 2

您可以将您的数字转换为字符串并像这样使用列表切片：

int(str(number)[:2])

输出：

>>> number = 1520
>>> int(str(number)[:2])
15

Answer 3

前两个答案都至少具有 O(n) 时间复杂度，并且字符串转换也具有 O(n) 空间复杂度。以下是恒定时间和空间的解决方案：

num // 10 ** (int(math.log(num, 10)) - 1)

功能：

import math

def first_n_digits(num, n):
    return num // 10 ** (int(math.log(num, 10)) - n + 1)

输出：

>>> first_n_digits(123456, 1)
1
>>> first_n_digits(123456, 2)
12
>>> first_n_digits(123456, 3)
123
>>> first_n_digits(123456, 4)
1234
>>> first_n_digits(123456, 5)
12345
>>> first_n_digits(123456, 6)
123456

如果您输入的数字可能比您想要的少，您将需要添加一些检查。

Answer 4

您可以使用正则表达式来测试匹配并捕获前两位数字：

import re

for i in range(1000):
    match = re.match(r'(1[56])', str(i))

    if match:
        print(i, 'begins with', match.group(1))

正则表达式 (1[56]) 匹配 1 后跟 5 或 6，并将结果存储在第一个捕获组中。

输出：

15 begins with 15
16 begins with 16
150 begins with 15
151 begins with 15
152 begins with 15
153 begins with 15
154 begins with 15
155 begins with 15
156 begins with 15
157 begins with 15
158 begins with 15
159 begins with 15
160 begins with 16
161 begins with 16
162 begins with 16
163 begins with 16
164 begins with 16
165 begins with 16
166 begins with 16
167 begins with 16
168 begins with 16
169 begins with 16

Answer 5

Dom's solution 实际上对 1000、1000000 等数字有问题。
这是因为

>>> math.log(1000, 10)  
2.9999999999999996

（结果 但是使用 math.log10() 就可以了。 (python 3.8)

>>> math.log10(1000)
3.0

也许函数 log 和 log10 有不同的计算算法。

固定功能是

import math

def first_n_digits(num, n):
    return num // 10 ** (int(math.log10(num)) - n + 1)

相关：Python math.log and math.log10 giving different results
（我可能应该对 dom's answer 发表评论，但我没有足够的声誉）