如何在二进制转换程序中的Python中每4位后面创建一个空格

Question

import collections
from collections import Counter
d = int(input('Pick a number to convert to Binary: '))

def convert(n, Counter = 0):
    Counter = 0
    if n > 1:
        convert(n//2)
    print(n % 2, end = '')
    
    
print("Your number in Binary is")
convert(d)

我不知道如何在输出中每 4 位或每 4 个数字创建一个空格。我尝试使用计数器、for 循环以及几乎所有我能想到的东西。我只是想知道我怎样才能做到这一点。任何帮助将不胜感激，我只是迷路了。

Answer 1

尝试这样做：

def convert(n, counter = None):
    if not counter:
        counter = 0
    counter += 1
    if n > 1:
        convert(n//2, counter)
    if counter % 4 == 0:
        print(" ", end="")
    print(n % 2, end = '')
    
    
    
print("Your number in Binary is")
convert(d)
print("")

3456 的输出作为输入：

Your number in Binary is
 1101 1000 0000

Answer 2

我也想解决这个问题。

在阅读了您的请求和其他答案后，我有动力在我的上下文中解决它。期望这是一个改进。

我得到了灵感，然后基于有问题的代码和之前的答案：

https://stackoverflow.com/a/69776706/26970467

这是从中得出的解决方案，也满足了我在这段代码下面概述的愿望：

# Created code is made for Python 3.12

# blank line to start the output lower in the console window
print("")
# Define variables
# get number, define and initialize d variable in the global scope
d = int(input('Pick a number to convert to Binary: '))


# Define function convert
def convert(n, counter=None, results=None, store_n=None):
    if not store_n:
        # initialize the store_n variable as the input n passed into the function.
        store_n = n
    if not results:
        # initialize the results variable as a list.
        # this allows the recursion to persist till function completes.
        results = []
    if not counter:
        # initialize the counter to zero.
        counter = 0
    # check if it is time to add a space to the result list. If so, do it.
    # if counter is not the initial value proceed.
    if counter != 0:
        # if counter is a multiple of 4, with remainder of zero, add space to results list.
        if counter % 4 == 0:
            results.append(" ")
    # increment counter by one
    counter += 1
    # put result of this recursive step into results list.
    # result is mod 2 of the current n in this recursive step.
    # the remainder of what n divided by 2 is. Remainder is either a 1 or a 0.
    results.append(str(n % 2))
    # check if this is a large number that requires more than 1 recursive step. if so, proceed.
    if n > 1:
        # pass next recursive step of n // 2 to function.
        # The floor division operator here returns n divided by 2 without any remainder.
        # also, passing current counter value, results and store_n static values into the recursive step.
        convert(n // 2, counter, results, store_n)
    # now we have all the results in a list but in reverse due to the recursion stepping.
    # we will reverse the sequence of collected results and make them a new string to
    # prevent the recursion from breaking the format.
    formated_results = str(''.join(list(reversed(results))))
    # get the length of the formated results string to determine how many digits exist in the result.
    # gather information to test and add zeros to the beginning of the formated_results string.
    # frl is a synonym for formated results length.
    frl = len(formated_results)
    # calculate the number of preceding zeros that are missing from the sequence of characters.
    missing_zeros = 5-((frl - 4) % 5)
    # calculate the number of digits that exist to the left of the first space.
    number_of_characters_beyond_the_seperator_space = ((frl - 4) % 5)-1
    # test and add zeros to the beginning of the formated_results string if needed.
    if number_of_characters_beyond_the_seperator_space > 0:
        formated_results = ("0" * missing_zeros) + formated_results
    return formated_results


# Main code:
# blank line
print("")
# give result
print("Your number", d, "in Binary is", convert(d))

对于我的用例情况，我喜欢在输出中调用更多约束：

• 我只需要在每组四位之间有一个空格分隔符。
• 如果第一个空格分隔符左侧有一系列少于 4 位的数字，我想将该字符集填充到 4 位。例如数字 16 作为输入。输出将为 0001 0000。
• 此外，在处理结果时，我不希望该系列以空格分隔符开头。
• 我想要一个可以返回结果的函数。

此外，您还会看到我对代码进行了注释，以便更好地了解代码的工作原理。

希望这对某人有帮助。