Python 中的二进制字符串问题

发布于 2024-12-07 16:09:43 字数 1762 浏览 1 评论 0 原文

出于某种原因,我花了很长时间才弄清楚如何在 Python 中做到这一点。 我试图在字符串变量中表示一个二进制字符串,而我想要它拥有的只是

0010111010

但是,无论我如何尝试将其格式化为字符串,Python总是会砍掉前导零,这让我很头疼试图解析它。

我希望这个问题会有有帮助,但实际上并没有...

有没有办法强制Python停止自动将我的字符串转换为整数?

我已经尝试过以下操作:

val = ""
if (random.random() > 0.50):
  val = val + "1"
else
  val = val + "0"

并且

val = ""
if (random.random() > 0.50):
  val = val + "%d" % (1)
else:
  val = val + "%d" % (0)

我之前已将其插入一个数组中,但在将该数组插入另一个数组时遇到了问题,所以我认为将其解析为字符串会更容易。

关于如何恢复我的前导零有什么想法吗?如果有帮助的话,该字符串应该是 10 位的固定长度。


编辑:

代码:

  def create_string(x):
    for i in xrange(10):        # 10 random populations
      for j in xrange(int(x)):  # population size
        v = ''.join(choice(('0','1')) for _ in range(10))
        arr[i][j] = v

    return arr

  a = create_string(5)
  print a

希望我看到的输出能够告诉您为什么我遇到问题:

[[  10000100 1100000001  101010110     111011   11010111]
 [1001111000 1011011100 1110110111  111011001   10101000]
 [ 110010001 1011010111 1100111000 1011100011 1000100001]
 [  10011010 1000011001 1111111010   11100110  110010101]
 [1101010000 1010110101  110011000 1100001001 1010100011]
 [  10001010 1100000001 1110010000   10110000   11011010]
 [    111011 1000111010    1100101 1101110001  110110000]
 [ 110100100 1100000000 1010101001   11010000 1000011011]
 [1110101110 1100010101 1110001110   10011111  101101100]
 [  11100010 1111001010  100011101    1101010 1110001011]]

这里的问题不仅与打印有关,我还需要能够在每个元素的基础上操作它们。因此,如果我去处理第一个元素,那么它会返回 1,而不是 0(在第一个元素上)。

For some reason I'm having a heck of a time figuring out how to do this in Python.
I am trying to represent a binary string in a string variable, and all I want it to have is

0010111010

However, no matter how I try to format it as a string, Python always chops off the leading zeroes, which is giving me a headache in trying to parse it out.

I'd hoped this question would have helped, but it doesn't really...

Is there a way to force Python to stop auto-converting my string to an integer?

I have tried the following:

val = ""
if (random.random() > 0.50):
  val = val + "1"
else
  val = val + "0"

and

val = ""
if (random.random() > 0.50):
  val = val + "%d" % (1)
else:
  val = val + "%d" % (0)

I had stuck it into an array previously, but ran into issues inserting that array into another array, so I figured it would just be easier to parse it as a string.

Any thoughts on how to get my leading zeroes back? The string is supposed to be a fixed length of 10 bits if that helps.


Edit:

The code:

  def create_string(x):
    for i in xrange(10):        # 10 random populations
      for j in xrange(int(x)):  # population size
        v = ''.join(choice(('0','1')) for _ in range(10))
        arr[i][j] = v

    return arr

  a = create_string(5)
  print a

Hopefully the output I'm seeing will show you why I'm having issues:

[[  10000100 1100000001  101010110     111011   11010111]
 [1001111000 1011011100 1110110111  111011001   10101000]
 [ 110010001 1011010111 1100111000 1011100011 1000100001]
 [  10011010 1000011001 1111111010   11100110  110010101]
 [1101010000 1010110101  110011000 1100001001 1010100011]
 [  10001010 1100000001 1110010000   10110000   11011010]
 [    111011 1000111010    1100101 1101110001  110110000]
 [ 110100100 1100000000 1010101001   11010000 1000011011]
 [1110101110 1100010101 1110001110   10011111  101101100]
 [  11100010 1111001010  100011101    1101010 1110001011]]

The issue here isn't only with printing, I also need to be able to manipulate them on a per-element basis. So if I go to play with the first element, then it returns a 1, not a 0 (on the first element).

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(2

ゃ人海孤独症 2024-12-14 16:09:43

如果我理解正确,你可以这样做:

a = 0b0010111010
'{:010b}'.format(a)

#The output is: '0010111010'

Python 2.7

它使用字符串格式方法

如果您想表示带有前导零的二进制字符串,这就是答案。

如果您只是尝试用二进制文件生成随机字符串,您可以这样做:

from random import choice
''.join(choice(('0','1')) for _ in range(10))

更新

不更新您的更新。
我编写了一个与您的代码相比具有不同输出的代码:

from random import choice
from pprint import pprint

arr = []

def create_string(x):
    for i in xrange(10):       # 10 random populations
        arr.append([])
        for j in xrange(x):  # population size
            v = ''.join(choice(('0','1')) for _ in range(10))
            arr[-1].append(v)
    return arr

a = create_string(5)
pprint(a)

输出是:

[['1011010000', '1001000010', '0110101100', '0101110111', '1101001001'],
 ['0010000011', '1010011101', '1000110001', '0111101011', '1100001111'],
 ['0011110011', '0010101101', '0000000100', '1000010010', '1101001000'],
 ['1110101111', '1011111001', '0101100110', '0100100111', '1010010011'],
 ['0100010100', '0001110110', '1110111110', '0111110000', '0000001010'],
 ['1011001011', '0011101111', '1100110011', '1100011001', '1010100011'],
 ['0110011011', '0001001001', '1111010101', '1110010010', '0100011000'],
 ['1010011000', '0010111110', '0011101100', '1111011010', '1011101110'],
 ['1110110011', '1110111100', '0011000101', '1100000000', '0100010001'],
 ['0100001110', '1011000111', '0101110100', '0011100111', '1110110010']]

这是您正在寻找的吗?

If I understood you right, you could do it this way:

a = 0b0010111010
'{:010b}'.format(a)

#The output is: '0010111010'

Python 2.7

It uses string format method.

This is the answer if you want to represent the binary string with leading zeros.

If you are just trying to generate a random string with a binary you could do it this way:

from random import choice
''.join(choice(('0','1')) for _ in range(10))

Update

Unswering your update.
I made a code which has a different output if compared to yours:

from random import choice
from pprint import pprint

arr = []

def create_string(x):
    for i in xrange(10):       # 10 random populations
        arr.append([])
        for j in xrange(x):  # population size
            v = ''.join(choice(('0','1')) for _ in range(10))
            arr[-1].append(v)
    return arr

a = create_string(5)
pprint(a)

The output is:

[['1011010000', '1001000010', '0110101100', '0101110111', '1101001001'],
 ['0010000011', '1010011101', '1000110001', '0111101011', '1100001111'],
 ['0011110011', '0010101101', '0000000100', '1000010010', '1101001000'],
 ['1110101111', '1011111001', '0101100110', '0100100111', '1010010011'],
 ['0100010100', '0001110110', '1110111110', '0111110000', '0000001010'],
 ['1011001011', '0011101111', '1100110011', '1100011001', '1010100011'],
 ['0110011011', '0001001001', '1111010101', '1110010010', '0100011000'],
 ['1010011000', '0010111110', '0011101100', '1111011010', '1011101110'],
 ['1110110011', '1110111100', '0011000101', '1100000000', '0100010001'],
 ['0100001110', '1011000111', '0101110100', '0011100111', '1110110010']]

Is this what you are looking for?

薄荷梦 2024-12-14 16:09:43

下面怎么样:

In [30]: ''.join('1' if random.random() > 0.50 else '0' for i in xrange(10))
Out[30]: '0000110111'

这给出了一个十个字符的二进制字符串;没有删除前导零。

如果您不需要改变数字概率(上面的 0.50),一个稍微简洁的版本是:

In [39]: ''.join(random.choice('01') for i in xrange(10))
Out[39]: '0001101001'

How about the following:

In [30]: ''.join('1' if random.random() > 0.50 else '0' for i in xrange(10))
Out[30]: '0000110111'

This gives a ten-character binary string; there's no chopping off of leading zeroes.

If you don't need to vary digit probability (the 0.50 above), a slightly more concise version is:

In [39]: ''.join(random.choice('01') for i in xrange(10))
Out[39]: '0001101001'
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文