对“/”类型字符的二进制搜索

Question

在二进制搜索中，我遇到了这个问题，无法将字符'/‘识别为字符串或int。因此，比较几乎是不可能的。我很好奇这个问题的想法是什么。如何对python中的url执行二进制搜索？

url1 = "https://diversity.google"
url2 = "https://www.aboutamazon.com/workplace/diversity-inclusion"
url3 = "https://www.indeed.com/q-Diversity-jobs.html?vjk=ba073b4704d48c67"
url4 = "https://careers.linkedin.com/diversity-and-inclusion"
url5 = "https://github.com/about/diversity"
url6 = "https://www.apple.com/diversity/"
url7 ="https://www.samsung.com/us/about-us/diversity-and-inclusion/"
url8 = "https://diversity.fb.com"
url9 ="instagram:none"
url10 = "https://careers.twitter.com/en/diversity.html"


data = [url1,url2,url3,url4,url5,url6,url7,url8,url9,url10]

print(data)
def binary_search(i, data, low, high, x):
  mid = (high+low) / 2

  if data[i][mid] > x:
    return binary_search(data, low, mid-1, x)
    
  elif data[i] [mid] < x:
      return binary_search(data, mid+1, high, x)

  else:
    return mid


error:

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-30-fb856fb84489> in <module>()
      9   high = [i][-1]
     10 
---> 11   foundOne = binary_search(i,data,low,high,x)
     12 
     13   if (foundOne == x):

<ipython-input-29-739df621f0c5> in binary_search(i, data, low, high, x)
     20 print(data)
     21 def binary_search(i, data, low, high, x):
---> 22   mid = (high+low) / 2
     23 
     24   if data[i][mid] > x:

TypeError: unsupported operand type(s) for /: 'str' and 'int'

(update)******

这就是我称之为it.Apologies的地方，因为它不包括它。

x = '/'
 



counterData = []

counter = 0
for i in data:
  low = [i][0]
  high = [i][-1]
  
  foundOne = binary_search(i,data,low,high,x)

  
  if (foundOne == x):
    counter += 1
  else:
    counter += 0
  
counterData.append(counter)

Answer 1

这是一个有效的版本。您可以看到它依次搜索每个条目，然后打印索引。

当我进行搜索时，我通常会把被搜索的东西称为“干草堆”，而我正在寻找的东西叫做“针”。

另外，您的函数不处理在干草堆中找不到针头的情况。

url1 = "https://diversity.google"
url2 = "https://www.aboutamazon.com/workplace/diversity-inclusion"
url3 = "https://www.indeed.com/q-Diversity-jobs.html?vjk=ba073b4704d48c67"
url4 = "https://careers.linkedin.com/diversity-and-inclusion"
url5 = "https://github.com/about/diversity"
url6 = "https://www.apple.com/diversity/"
url7 = "https://www.samsung.com/us/about-us/diversity-and-inclusion/"
url8 = "https://diversity.fb.com"
url9 = "instagram:none"
url10 = "https://careers.twitter.com/en/diversity.html"


data = [url1,url2,url3,url4,url5,url6,url7,url8,url9,url10]
data.sort()

def binary_search(haystack, low, high, needle):
    if high < low:
        return -1

    mid = (high+low) // 2

    if haystack[mid] > needle:
        return binary_search(haystack, low, mid-1, needle)
    elif haystack[mid] < needle:
        return binary_search(haystack, mid+1, high, needle)
    else:
        return mid

for i in data:
    foundOne = binary_search(data, 0, len(data)-1, i)
    print(i, foundOne)

输出：

https://careers.linkedin.com/diversity-and-inclusion 0
https://careers.twitter.com/en/diversity.html 1
https://diversity.fb.com 2
https://diversity.google 3
https://github.com/about/diversity 4
https://www.aboutamazon.com/workplace/diversity-inclusion 5
https://www.apple.com/diversity/ 6
https://www.indeed.com/q-Diversity-jobs.html?vjk=ba073b4704d48c67 7
https://www.samsung.com/us/about-us/diversity-and-inclusion/ 8
instagram:none 9