让二进制搜索在 Python 中工作

Question

我正在尝试让二进制搜索在 Python 中工作。我有一个庞大的、排序的密码列表。计划是从用户那里获取密码输入并查看它是否在列表中。由于列表的大小，我决定实现二进制搜索。

这是我的代码：

Found = False
Password = user_input("Enter a password: ")


with io.open('final.txt', encoding='latin-1') as myfile:

    data = myfile.readlines()
    low = 0
    high = (int(len(data))+1)
    while (low < high) and not Found:

        mid = int((low+high)/2)

        if data[mid] == Password:
            Found = True
            break
        elif Password < str(data[mid]):
            high = mid - 1
        elif Password > str(data[mid]):
            low = mid + 1

我猜是因为字符串比较？有任何想法吗？二进制搜索永远不会返回 true，即使我明确搜索我知道在列表中的内容。

我使用此代码对密码列表进行排序。

import io

with io.open('result.txt', encoding='latin-1') as myfile:
    data = myfile.readlines()

def partition(data, start, end):
    pivot = data[end]                          # Partition around the last value
    bottom = start-1                           # Start outside the area to be partitioned
    top = end                                  # Ditto

    done = 0
    while not done:                            # Until all elements are partitioned...

        while not done:                        # Until we find an out of place element...
            bottom = bottom+1                  # ... move the bottom up.

            if bottom == top:                  # If we hit the top...
                done = 1                       # ... we are done.
                break

            if data[bottom] > pivot:           # Is the bottom out of place?
                data[top] = data[bottom]       # Then put it at the top...
                break                          # ... and start searching from the top.

        while not done:                        # Until we find an out of place element...
            top = top-1                        # ... move the top down.

            if top == bottom:                  # If we hit the bottom...
                done = 1                       # ... we are done.
                break

            if data[top] < pivot:              # Is the top out of place?
                data[bottom] = data[top]       # Then put it at the bottom...
                break                          # ...and start searching from the bottom.

    data[top] = pivot                          # Put the pivot in its place.
    return top                                 # Return the split point


def quicksort(data, start, end):
    if start < end:                            # If there are two or more elements...
        split = partition(data, start, end)    # ... partition the sublist...
        quicksort(data, start, split-1)
        quicksort(data, split+1, end)


quicksort(data, 0, (int(len(data))-1))

with io.open('final.txt', 'w', encoding='latin-1') as f:
    for s in data:
        f.write(s)

排序后的列表如下所示：空格，然后是符号，然后是数字，然后是大写字母（按字母顺序排序），然后是普通字母（按字母顺序排序）。

Answer 1

由于您设置low 和high 的方式，您正在跳过部分列表。正因为如此，low == high发生在更新后检查前，导致你过早跳出循环。

有两种简单的解决方案：

要么..

• 设置high = mid 或low = mid 而不是mid -/+ 1，触发额外的迭代，

或者..

• 检查循环后是否high == low and data[low] == Password 终止，因为您可能仍会在那里找到Password。

Answer 2

有两个问题。

1. 您的二分搜索算法错误。

重复条件应该是

while (low <= high)

或者你找不到第一个和最后一个元素。

1. readlines() 将读取 \n 但 user_input() 不会。

这会导致`Password` == `Password\n' 永远为假。

Answer 3

这是二分查找的例子

def binarySearch(alist, item):
        first = 0
        last = len(alist)-1
        found = False

        while first<=last and not found:
            midpoint = (first + last)//2
            if alist[midpoint] == item:
                found = True
            else:
                if item < alist[midpoint]:
                    last = midpoint-1
                else:
                    first = midpoint+1

        return found

mylist1 = [0, 1, 2, 8, 9, 17, 19, 32, 42,]
print(binarySearch(mylist1, 3))
print(binarySearch(mylist1, 13))

mylist2 = [0, 1, 2, 8, 9, 17, 19, 32, 42, 99]
print(binarySearch(mylist2, 2))
print(binarySearch(mylist2, 42))

我明白了

False
False
True
True

是的，正如 Eamon 指出的那样，我确信在调用 readlines 后，您需要在每个密码的末尾添加换行符。

Answer 4

您可能在调用readlines 后每个密码的末尾都有一个换行符，使用rstrip() 将其删除

    Found = False
    Password = user_input("Enter a password: ")


    with io.open('final.txt', encoding='latin-1') as myfile:

        data = myfile.readlines()
        low = 0
        high = len(data)-1   #no need to cast to int, should be len()-1
        while (low <= high) and not Found:  #less than or equal to

            mid = int((low+high)/2)

            if data[mid].rstrip() == Password:   #Remove newline character before compare
                Found = True
                break
            elif Password < str(data[mid]):
                high = mid - 1
            elif Password > str(data[mid]):
                low = mid + 1

Answer 5

如果你只想在你的列表中搜索密码，那么在你的代码中

data = myfile.readlines()

你已经把所有的密码都存入了内存。 因此，如果您只想检查给定密码是否存在于您的列表中，您可以使用直接检查

if Password in data:
     print "yes it is present in the list"
else:
    print "Not present in the list"

希望能有所帮助。

Answer 6

不要编写自己的二进制搜索，要正确处理它们有点棘手。请改用bisect 模块。

from bisect import bisect_left
def binary_search(lst, el):
   # returns lower bound of key `el` in list `lst`
   index = bisect_left(lst, el)
   # check that: (1) the lower bound is not at the end of the list and
   # (2) the element at the index matches `el`
   return index < len(lst) and lst[index] == el

用法：

test = ["abc", "def", "ghi"]
print(binary_search(test, "def")) # True
print(binary_search(test, "xyz")) # False