将数字的最后n位数转换为零

Question

What is the best way to replace the last digits of a number to zero keeping first three digits as it is, in python?

Example:

23456789 -> 23400000
112022 -> 112000
1111-> 1110
111 -> 111 (no conversion)

Answer 1

convert it to a string:

a = 23456789

a_str = str(a)

sol = a_str[0:3]+"0"*(len(a_str)-3)

print(sol)

returns: 23400000

Answer 2

round accepts a negative argument, which is quite useful, but it rounds the result, which may not be exactly what you want:

a = 23456789
round(a, -4)

outputs:

23460000

If you do not wish to round , you could use a small helper function, maybe like this one:

import math

def significant(num, signum):

    expo = 10**(int(math.log(num, 10)) - signum + 1)
    return expo * (num // expo)

significant(12345, 3)

outputs:

12300

Answer 3

Use round , but you have to be a bit more clever to get the functionality you want

def round_n(num, keep=3):
    return round(num, keep - len(str(num)))


round_n(23456789)
Out[72]: 23500000

round_n(112022)
Out[73]: 112000

round_n(1111)
Out[74]: 1110

round_n(111)
Out[75]: 111

---

nb this will only work with integers, not floats

---

If you just want to truncate, not round, then instead do

def truncate_int(n, keep=3):
   if n < 0:  # account for '-'
       keep += 1
   s = str(n)
   return int(s[:keep] + '0'*(len(s) - keep))

Answer 4

You can define a converter function using math.log10 , floor division and exponentials:

a = 23456789
b = 112022
c = 1111
d = 111

import math

def converter(x, k=3):
    dig = int(math.log10(x)) + 1
    n = dig - k
    return x // 10**n * 10**n

for val in [a, b, c, d]:
    print(val, '->', converter(val))

23456789 -> 23400000
112022 -> 112000
1111 -> 1110
111 -> 111

Answer 5

I would do this using floor division and multiplication with an appropriate multiplier. That way, we avoid problems with rounding. We also have to ensure that the multiplier is an integer. We can do that by using max to force m to be non-negative.

data = (23456789, 112022, 1111, 111, 6, 98, 987, 9876, 9876598765)

def zero_final(n, digits=3):
    m = 10 ** max(0, len(str(n)) - digits)
    return n // m * m

for n in data:
    print(n, zero_final(n))

output

23456789 23400000
112022 112000
1111 1110
111 111
6 6
98 98
987 987
9876 9870
9876598765 9870000000

Answer 6

Not sure it's the best way,

but I would use string manipulation and formatting.

number = 23456789
print int(str(number)[:3] + "%0{0}d".format(len(str(number)) - 3) % 0)

Answer 7

You can try using string operation as follows:

def number_cov(num):
    x = str(num)
    if len(x)>3:
        cov_num = int(x[0:3]+'0'*(len(x)-3))
    else:
        cov_num = num
    return cov_num

number_cov(11111)
11100

Answer 8

n - n%1000

For a positive integer n , n%1000 gives you the value of the last three digits. So subtracting that will give you a number ending in 000 .

>>> n = 12345
>>> n - n%1000
12000

If you want to do this for other numbers of digits, you can do it like this:

>>> n = 1234567
>>> r = 10**5 # I want to zero-out five digits
>>> n - n%r
1200000