python - BinarySearch 修改

Question

嗨，我正在修改二进制搜索，以计算 url 中“/”的数量，并根据该数量创建一个分数基础，并为其创建一个 pi 图。但是，当我尝试遍历字符串时遇到错误。如果您还可以帮助我找到一种更简单的方法来对列表中的 url 进行排序，而不是对每个 url 进行手动排序，然后将其组合到一个列表中，那就太棒了。谢谢！

import numpy as np

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"


#for some reason doing this in a list doesnt work
url1 = sorted(url1)
url1 = "".join(url1)

url2 = sorted(url2)
url2 = "".join(url2)

url3 = sorted(url3)
url3 = "".join(url3)

url4 = sorted(url4)
url4 = "".join(url4)

url5 = sorted(url5)
url5 = "".join(url5)

url6 = sorted(url6)
url6 = "".join(url6)

url7 = sorted(url7)
url7 = "".join(url7)

url8 = sorted(url8)
url8 = "".join(url8)

url9 = sorted(url9)
url9 = "".join(url9)

url10 = sorted(url10)
url10 = "".join(url10)




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

#for i in data:
  #data[i] = sorted(data[i])
  #data[i] = "".join(data[i])

def BinaryMod(data, low, high, x, counterArray):
  
  mid = (high+low) / 2
  for i in data:
    for j in len(data[i]):
      word = j
      high = len(word) - 1
      if word[mid] > x:
        end = mid - 1
        
      elif word[mid] < x:
        low = mid+1

      else:
        counter = counter + 1

    counterArray.append(counter)
 return counterArray

counterArray = np.array([])

counterArray = BinaryMod(data, 0, 0,  '/', counterArray)


IndexError                                Traceback (most recent call last)
<ipython-input-29-1293e7cb075f> in <module>()
      1 counterArray = np.array([])
      2 
----> 3 counterArray = BinaryMod(data, 0, 0,  '/', counterArray)

<ipython-input-28-4fe8d062745d> in BinaryMod(data, low, high, x, counterArray)
     62   mid = (high+low) / 2
     63   for i in data:
---> 64     for j in len(data[i]):
     65       word = j
     66       high = len(word) - 1

IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices

标签： pythondata-sciencebinary-search

Answer 1

import numpy as np

urls = [
    "https://diversity.google",
    "https://www.aboutamazon.com/workplace/diversity-inclusion",
    "https://www.indeed.com/q-Diversity-jobs.html?vjk=ba073b4704d48c67",
    "https://careers.linkedin.com/diversity-and-inclusion",
    "https://github.com/about/diversity",
    "https://www.apple.com/diversity/",
    "https://www.samsung.com/us/about-us/diversity-and-inclusion/",
    "https://diversity.fb.com",
    "instagram:none",
    "https://careers.twitter.com/en/diversity.html",
]


for url in urls:
    print( url, 'contains', url.count('/'), 'slashes' )

输出：

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