Как найти тысячи в питоне

num = int(input("Please give me a number: "))
print(num)
thou = int((num // 1000))
print(thou)
hun = int((num // 100))
print(hun)
ten =int((num // 10))
print(ten)
one = int((num // 1))
print(one)

I tried this but it does not work and I’m stuck.

You might want to try something like following:

def get_pos_nums(num):
    pos_nums = []
    while num != 0:
        pos_nums.append(num % 10)
        num = num // 10
    return pos_nums

And call this method as following.

>>> get_pos_nums(9876)
[6, 7, 8, 9]

The 0th index will contain the units, 1st index will contain tens, 2nd index will contain hundreds and so on…

This function will fail with negative numbers. I leave the handling of negative numbers for you to figure out as an exercise.

answered Jun 28, 2018 at 0:28

num = 1234

thousands = num // 1000
hundreds = (num % 1000) // 100
tens = (num % 100) // 10
units = (num % 10)

print(thousands, hundreds, tens, units)
# expected output: 1 2 3 4

“//” in Python stands for integer division. It largely removes the fractional part from the floating-point number and returns an integer

For example:

4/3 = 1.333333
4//3 = 1

answered Jun 8, 2021 at 2:19

You could try splitting the number using this function:

def get_place_values(n):
    return [int(value) * 10**place for place, value in enumerate(str(n)[::-1])]

For example:

get_place_values(342)
>>> [2, 40, 300]

Next, you could write a helper function:

def get_place_val_to_word(n):
    n_str = str(n)
    num_to_word = {
        "0": "ones",
        "1": "tens",
        "2": "hundreds",
        "3": "thousands"
    }
    return f"{n_str[0]} {num_to_word[str(n_str.count('0'))]}"

Then you can combine the two like so:

def print_place_values(n):
    for value in get_place_values(n):
        print(get_place_val_to_word(value))

For example:

num = int(input("Please give me a number: "))
# User enters 342
print_place_values(num)
>>> 2 ones
4 tens
3 hundreds

answered Jun 29, 2021 at 2:19

num=1234
digit_at_one_place=num%10
print(digit_at_one_place)
digits_at_tens_place=(num//10)%10
digits_at_hund_place=(num//100)%10
digits_at_thou_place=(num//1000)%10
print(digits_at_tens_place)
print(digits_at_hund_place)
print(digits_at_thou_place)

this does the job. it is simple to understand as well.

answered Nov 26, 2021 at 5:21

Please note that I took inspiration from the above answer by 6pack kid to get this code. All I added was a way to get the exact place value instead of just getting the digits segregated.

num = int(input("Enter Number: "))
c = 1
pos_nums = []
while num != 0:
    z = num % 10
    pos_nums.append(z *c)
    num = num // 10
    c = c*10
print(pos_nums)

Once you run this code, for the input of 12345 this is what will be the output:

Enter Number: 12345
[5, 40, 300, 2000, 10000]

This helped me in getting an answer to what I needed.

money = int(input("Enter amount: "))
thousand = int(money // 1000)
five_hundred = int(money % 1000 / 500)
two_hundred = int(money % 1000 % 500 / 200)
one_hundred = int(money % 1000 % 500 % 200 / 100)
fifty  = int(money % 1000 % 500 % 200 % 100 / 50)
twenty  = int(money % 1000 % 500 % 200 % 100 % 50 / 20)
ten  = int(money % 1000 % 500 % 200 % 100 % 50 % 20 / 10)
five  = int(money % 1000 % 500 % 200 % 100 % 50 % 20 % 10 / 5)
one = int(money % 1000 % 500 % 200 % 100 % 50 % 20 % 10 % 5 / 1)
if thousand >=1: 
  print ("P1000: " , thousand)
if five_hundred >= 1:
  print ("P500: " , five_hundred)
if two_hundred >= 1:
  print ("P200: " , two_hundred)
if one_hundred >= 1:
  print ("P100: " , one_hundred)
if fifty >= 1:
  print ("P50: " , fifty)
if twenty >= 1:
  print ("P20: " , twenty)
if ten >= 1:
  print ("P10: " , ten)
if five >= 1:
  print ("P5: " , five)
if one >= 1:
  print ("P1: " , one)

answered Feb 2, 2020 at 14:53

Quickest way:

num = str(input("Please give me a number: "))
print([int(i) for i in num[::-1]])

answered Jun 29, 2021 at 11:45

This will do it, doesn’t use strings at all and handles any integer passed for col sensibly.

def tenscol(num: int, col: int):
    ndigits = 1
    while (num % (10**ndigits)) != num:
        ndigits += 1
    x = min(max(1, col), ndigits)
    y = 10**max(0, x - 1)
    return int(((num % 10**x) - (num % y)) / y)

usage:

print(tenscol(9785,-1))
print(tenscol(9785,1))
print(tenscol(9785,2))
print(tenscol(9785,3))
print(tenscol(9785,4))
print(tenscol(9785,99))

Output:

5
5
8
7
9
9

answered Nov 6, 2021 at 21:56

def get_pos(num,unit):
    return int(abs(num)/unit)%10

So for “ones” unit is 1 while for “tens”, unit is 10 and so forth.

It can handle any digit and even negative numbers effectively.

So given the number 256, to get the digit in the tens position you do

get_pos(256,10)
>> 5

answered Nov 12, 2021 at 22:04

I had to do this on many values of an array, and it’s not always in base 10 (normal counting – your tens, hundreds, thousands, etc.). So the reference is slightly different: 1=1st place (1s), 2=2nd place (10s), 3=3rd place (100s), 4=4th place (1000s). So your vectorized solution:

import numpy as np
def get_place(array, place):
    return (array/10**(place-1)%10).astype(int)

Works fast and also works on arrays in different bases.

answered Apr 7, 2022 at 20:15

So I saw what another users answer was and I tried it out and it didn’t quite work, Here’s what I did to fix the problem. By the way I used this to find the tenth place of a number

# Getting an input from the user

input = int(input())

# Finding the tenth place of the number

print(int(input % 100) // 10)

answered Jun 28, 2021 at 20:38

задан 18 июл 2016 в 8:17

Если хочется сгруппировать тысячи в числе в соответствии с текущей локалью пользователя:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, '')
'ru_RU.UTF-8'
>>> print(locale.format('%d', 1234567, grouping=True))
1 234 567

В другой локали, разделитель может быть другим:

>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')
'en_US.UTF-8'
>>> print(locale.format('%d', 1234567, grouping=True))
1,234,567

Имена локалей и доступная функциональность могут зависеть от системы.

ответ дан 20 июл 2016 в 17:40

так можно любую маску сделать по количеству разрядов

import re

def numberjump_filter(string):
    string=str(string)
    #s = re.match("(d{0,})(d{3})(d{4})",string)
    s = re.match("(.*)(ddd)(dddd)",string)
    return s[1]+" "+s[2]+" "+s[3]

ответ дан 4 фев 2019 в 10:25

Ещё вариант для разнообразия:

n = str(1234567)[::-1]
print(' '.join(n[i:i+3] for i in range(0, len(n), 3))[::-1])

• делаем из числа строку
• переворачиваем её
• отрезаем по три символа
• объединяем их через пробелы
• ещё раз переворачиваем то, что получилось

Вывод:

1 234 567

ответ дан 19 окт 2021 в 13:28