檢查連續零的字串

我想制作一個回傳連續零數量的函式。

print(consecutive_zeros("0100100001"))

應該回傳 4

def consecutive_zeros(string):
    zeros = 0
    cache = 0
    for elem in range(len(string)):
        if string[elem] == "0":
            zeros  = 1
        elif string[elem] == "1":
            if zeros > cache:
                zeros = cache
                zeros = 0
        return cache

如果我給它一個帶數字的字串，我的 func 總是回傳 0

for elem in range(len(string)):
    if string[elem] == "0":
        zeros  = 1

這應該每 0 計數一次

elif string[elem] == "1":
        if zeros > cache:
            zeros = cache

這會在 1 出現時打破計數器，并檢查計數的連續零的數量是否比暫時保存的零更高

uj5u.com熱心網友回復：

我不是真正的 Python 人，但我只是在這里使用正則運算式。

txt = "0100100001"

# Find all sub strings with 2 or more consecutive zeros
matches = re.findall("00 ", txt)

# Compute string lengths
lengths = map(lambda x: len(x), matches)

# Find Longest
max_zeros = max(lengths);

print(max_zeros)

uj5u.com熱心網友回復：

這是您的代碼的修復：

def consecutive_zeros(string):
    zeros = 0
    cache = 0
    for elem in string:
        if elem == "0":
            zeros  = 1
            if zeros > cache:
                cache = zeros # this was inverted
        else:
                zeros = 0
    return cache # this was wrongly indented

consecutive_zeros("01001000010")

輸出： 4

我個人會使用itertools.groupby和理解：

from itertools import groupby 

s = '0100100001'

max(len(list(g)) for v,g in groupby(s) if v=='0')

輸出： 4

uj5u.com熱心網友回復：

您還需要在完成迭代后檢查當前計數，以防最大連續出現在最后一個序列中。

def consecutive_zeros(string):
    curr = 0
    max_consecutive_zeros = 0
    for ch in string:
        if ch == '0':
            curr  = 1
            continue
        else:
            if curr >= max_consecutive_zeros:
                max_consecutive_zeros = curr
            curr = 0
    if curr >= max_consecutive_zeros:
        max_consecutive_zeros = curr
    return maxmax_consecutive_zeros    
max_consecutive_zeros = consecutive_zeros("0100100001")

print(max_consecutive_zeros)
>> 4

uj5u.com熱心網友回復：

請參閱@Nick B 以獲得出色（且快速！）的答案。

但是，如果您想使用您的解決方案；

1. 您的函式在第一個回圈后回傳 - 移動您的函式return，使其與您的for回圈對齊

def consecutive_zeros(string):
    zeros = 0
    cache = 0
    for elem in range(len(string)):
        if string[elem] == "0":
            zeros  = 1
        elif string[elem] == "1":
            if zeros > cache:
                zeros = cache
                zeros = 0
    return cache #Aligned it such that data is returned at last

2. 您的函式回傳cache永遠不會更新；你需要更新

def consecutive_zeros(string):
    zeros = 0
    cache = 0
    for elem in range(len(string)):
        if string[elem] == "0":
            zeros  = 1
        elif string[elem] == "1":
            if zeros > cache:
                cache = zeros #'zeros = cache' before
                zeros = 0
    return cache #Aligned it such that data is returned at last

3. 您的代碼不會重置zeros- 僅當zeros>cache. 移到zeros=0外面，if zeros > cache但仍然在elif string[elem] == "1"

def consecutive_zeros(string):
    zeros = 0
    cache = 0
    for elem in range(len(string)):
        if string[elem] == "0":
            zeros  = 1
        elif string[elem] == "1":
            if zeros > cache:
                cache = zeros #'zeros = cache' before
            zeros = 0 #Move it such that it's updated whenever we reset our counter
    return cache #Aligned it such that data is returned at last

然后它起作用了

4. 注意，使用for elem in string代替for elem in range(len(string))

uj5u.com熱心網友回復：

如果您的字串只包含0和1's，您可以：

data = "0100100001"
print(max(len(x) for x in data.split('1')))

列印：

4

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