Python搜索字典键用于搜索输入 [英] Python search dictionary keys for search input

问题描述

所以这是我的问题：

我想搜索一个字典，看看是否有任何键包含用户输入的关键字。例如，用户搜索John。

  elif选项== 3：
 count = 0 
 found = None 
 search_key = input（你想要搜索什么？）.lower（）
为key，val在telephone_directory.items（）中：#takes电话目录中的每个元素
如果search_key在密钥中：#checks是否包含search_key 
如果发现是无：
 found = val 
 count = 1 
如果发现不是没有：
 print（）
 print（找不到一个匹配请更具体一些。）
 print（）
 count = 2 
 break 
 
如果发现是无：
 print（对不起，+ str（search_key）+未找到。）
 print（）
 function_options（）#redirects如果找到，则
 
不是无，计数< 2：
 print（str（search_key）+在目录中找到。）
 print（这里是文件+ str（search_key）+：）
打印（str（search_key）+：++ phone_directory [search_key]）
 print（）
 function_options（）#redirects back 
  pre> 
 
 
所以这是我现在所在的地方。无论搜索可能是什么，即使是整个键，它返回未找到。我做错了什么？
解决方案
你需要做几个选择;允许多个比赛，只找到第一场比赛，或者只允许最多一场比赛。
 
 
 
要找到第一场比赛，请使用 next（） ：
  match = next（val for key，val in telephone_directory.items（）if search_key in关键）
  
如果没有，这将引发 StopIteration 比赛被发现;返回默认值或捕获异常：
 ＃默认为无
 match = next（（val为key，val in my_dict.items（）if search_key in key），None）
 
 try：
 match = next（val for key，val in telephone_directory.items（）if search_key in key ）
除了StopIteration：
 print（Not found）
  
这些版本只会循环查找字典项，直到发现匹配，然后停止; $ / code code code code code code code code code $ phone_directory.items（）：
如果search_key在密钥中：
 print（找到一个匹配！{}。format（val））
 break 
 else：
 print（Nothing found）
  

请注意 else 块只有当循环的被允许完成时，才被调用，并且没有被 break 语句中断。 p>



要查找所有匹配键，请使用列表理解：

  matches = [val for key，val in phone_directory.items（）if search_key in key] 
  

最后，只允许一个匹配，高效地在同一个迭代器上使用两个 next（）调用，如果发现第二个匹配则引发错误

$ b

  def find_one_match（d，search_key）：
d = iter（d.items（））
尝试：
 match = next（val为键，val in d，如果search_key在键中）
除了StopIteration：
 raise ValueError（'Not found'）
 
 if $（$）
 raise ValueError（'多于一个匹配'）
 
返回匹配
  

适应这个在的循环方法中，只有在找到第二个项目时才需要破解：

  found = None 
为key，val在telephone_directory.items（）中：
如果search_key在key中：
如果找到则为None：
 found = val 
其他：
打印（找到多个匹配，请更具体）
 break 
 else：
如果发现是无：
打印（没有找到，请重新搜索）
其他：
打印（匹配发现！ {}。format（found））
  

您的版本无效，因为打印未找到'for 每个密钥不匹配，您只能知道，当您迭代了字典中的所有密钥时，您才能直到最终不匹配。 p>


So here's my question:

I want to search a dictionary to see if any key contains a user-inputted keyword. For example, the user searches for John.

elif option == 3:
        count = 0
        found = None
        search_key = input("What do you want to search for? ").lower()
        for key, val in telephone_directory.items(): #takes each element in telephone directory
            if search_key in key: #checks if it contains search_key
                if found is None:
                    found = val
                    count = 1
                if found is not None:
                    print(" ")
                    print("More than one match found. Please be more specific.")
                    print(" ")
                    count = 2
                    break

            if found is None:
                print("Sorry, " + str(search_key) + " was not found.")
                print(" ")
                function_options() #redirects back

            if found is not None and count < 2:
                print(str(search_key) + " was found in the directory.")
                print("Here is the file on " + str(search_key) + ":")
                print(str(search_key) + ":" + " " + telephone_directory[search_key])
                print(" ")
                function_options() #redirects back  

So this is where I am right now. Whatever the search may be, even if it is the entire key, it returns "was not found". What am I doing wrong?

解决方案

You need to make a few choices; allow multiple matches, find only the first match, or only allow for at most one match.

To find the first match, use next():

match = next(val for key, val in telephone_directory.items() if search_key in key)

This will raise StopIteration if no match was found; return a default instead or catch the exception:

# Default to `None`
match = next((val for key, val in my_dict.items() if search_key in key), None)

try:
    match = next(val for key, val in telephone_directory.items() if search_key in key)
except StopIteration:
    print("Not found")

These versions will only loop over the dictionary items until a match is found, then stop; the full for loop equivalent would be:

for key, val in telephone_directory.items():
    if search_key in key:
        print("Found a match! {}".format(val))
        break
else:
    print("Nothing found")

Note the else block; it is only called when the for loop was allowed to complete, and was not interrupted by a break statement.

To find all matching keys, use can use a list comprehension:

matches = [val for key, val in telephone_directory.items() if search_key in key]

Finally, to allow for only one match, efficiently, use two next() calls on the same iterator, and raise an error if a second match is found:

def find_one_match(d, search_key):
     d = iter(d.items())
     try:
         match = next(val for key, val in d if search_key in key)
     except StopIteration:
         raise ValueError('Not found')    

     if next((val for key, val in d if search_key in key), None) is not None:
         raise ValueError('More than one match')

     return match

Adapting that to the for loop approach again, would require you to break only if a second item is found:

found = None
for key, val in telephone_directory.items():
    if search_key in key:
        if found is None:
            found = val
        else:
            print("Found more than one match, please be more specific")
            break
else:
    if found is None:
        print("Nothing found, please search again")
    else:
        print("Match found! {}".format(found))

Your version doesn't work because you print 'not found' for each and every key that doesn't match. You can only know that you didn't match a key until the very end when you've iterated over all the keys in your dictionary.

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