如何将代码修改为所需的格式？

Question

def rle_encode(data):
    encoding = ''
    prev_char = ''
    count = 1

    if not data: return ''

    for char in data:
        # If the prev and current characters
        # don't match...
        if char != prev_char:
            # ...then add the count and character
            # to our encoding
            if prev_char:
                encoding += str(count) + prev_char
            count = 1
            prev_char = char
        else:
            # Or increment our counter
            # if the characters do match
            count += 1
    else:
        # Finish off the encoding
        encoding += str(count) + prev_char
        return encoding

encoded_val = rle_encode(“aaaaBBbbCooa”)
print(encoded_val)

对于上面的字符串，我得到 output 为：“4a2B2b1C2o1a”

但预期的 output 是：“4a2B2bC2oa”

这意味着对于单个字符，我不需要输入 ist 编号，只需输入字符即可。

如果有人告诉我在哪里更改代码以获得 output 的预期格式，那就太好了。

Answer 1

您可以通过使用itertools.groupby来实现。 如果组的长度大于 1，则添加计数，否则只需添加字符本身。

from itertools import groupby

def rle_encode(data):
    out = ''
    for key, group in groupby(data):
        run = len(list(group))
        if run > 1:
            out += str(run) + key
        else:
            out += key
    return out

例子

>>> rle_encode("aaaaBBbbCooa")
'4a2B2bC2oa'

Answer 2

这是您的固定代码。

def rle_encode(data):
    encoding = ''
    prev_char = ''
    count = 1

    if not data: return ''

    for char in data:
        # If the prev and current characters
        # don't match...
        if char != prev_char:
            # ...then add the count and character
            # to our encoding
            if prev_char:
                if count > 1:
                    encoding += str(count) + prev_char
                else:
                    encoding += prev_char
            count = 1
            prev_char = char
        else:
            # Or increment our counter
            # if the characters do match
            count += 1
    else:
        # Finish off the encoding
        if count > 1:
            encoding += str(count) + prev_char
        else:
            encoding += prev_char
        return encoding

Answer 3

显然，@Cory 的解决方案更好，它甚至大大简化了您的 function。 但是如果你仍然想要你的解决方案，我认为这就像删除'1's一样简单

def rle_encode(data):
    encoding = ''
    prev_char = ''
    count = 1

    if not data: return ''

    for char in data:
        # If the prev and current characters
        # don't match...
        if char != prev_char:
            # ...then add the count and character
            # to our encoding
            if prev_char:
                encoding += str(count) + prev_char
            count = 1
            prev_char = char
        else:
            # Or increment our counter
            # if the characters do match
            count += 1
    # Finish off the encoding
    encoding += str(count) + prev_char
    return encoding.replace('1', '')

encoded_val = rle_encode('aaaaBBbbCooa')
print(encoded_val)

顺便说一句，我认为您的最后一个“其他”是不必要的。

Answer 4

    def rle_encode(data):
    encoding = ''
    prev_char = ''
    count = 1

    if not data: return ''

    for char in data:
        # If the prev and current characters
        # don't match...
        print(char," ", count)
        if char != prev_char:
            # ...then add the count and character
            # to our encoding
            if prev_char:
                if count != 1:
                    encoding += str(count) + prev_char
                else:
                    encoding += prev_char

            count = 1
            prev_char = char
        else:
            # Or increment our counter
            # if the characters do match
            count += 1
    else:
        # Finish off the encoding
        if count != 1:
            encoding += str(count) + prev_char
        else:
            encoding += prev_char
        return encoding

encoded_val = rle_encode('aaaaBBbbCooa')
print(encoded_val)


Hope this will help you.