如何在python中不使用内置方法查找关键字前后的单词
我想使用内置方法查找关键字wihtout前后的单词 代码使用如何在python中不使用内置方法查找关键字前后的单词,python,Python,我想使用内置方法查找关键字wihtout前后的单词 代码使用分区方法工作正常 mystring = "This is python class" keyword = ' is' before_keyword, keyword, after_keyword = mystring.partition(keyword) l = [] l.append(before_keyword) l.append(keyword) l.append(after_keyword) print(l
分区mystring = "This is python class"
keyword = ' is'
before_keyword, keyword, after_keyword = mystring.partition(keyword)
l = []
l.append(before_keyword)
l.append(keyword)
l.append(after_keyword)
print(l)
['This'，'is'，'python class']“这是要测试的hello”'is'您可以尝试使用
插入和拆分，如下所示：
mystring = mystring.split(keyword)
mystring.insert(1, keyword)
print(mystring)
mystring = "This is python class"
keyword = ' as'
l = []
x = mystring.find(keyword)
if x != -1:
l.append(mystring[:x])
l.append(keyword)
l.append(mystring[x+len(keyword):])
print (l)
else:
print (keyword, 'is not found in', mystring)
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
编辑：
要使代码适用于各种字符串，可以使用以下大量代码：
mystring = "This is python class and it is a unicorn"
keyword = 'is'
newlist = []
if keyword in mystring:
string = ''
for i in mystring.split():
if i == keyword:
newlist.append(string)
newlist.append(i)
string = ''
else:
string += i + ' '
newlist.append(string)
newlist = [i.strip() for i in newlist if i]
print(newlist)
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
假设您不想使用内置方法，您可以创建自己的分区
或非分区
：p
inp = "This hello is to test"
def not_partition(inp='', part_at=''):
inp = inp.strip().split()
ret = ['']
for i,v in enumerate(inp):
if v==part_at:
ret = ret+[part_at, '']
else:
ret[-1] = (ret[-1]+' '+v).strip()
if ret[-1]=='': del ret[-1]
return ret
not_partition(inp, 'is'), not_partition(inp, 'test')
您还可以创建一个更高效的递归not\u分区
：D
def not_partition(inp='', part_at=''):
inp = inp.strip().split()
i=0
while i<len(inp):
if inp[i]==part_at:
return [' '.join(inp[:i]), part_at]+not_partition(' '.join(inp[i+1:]),part_at)
else:
i+=1
return [' '.join(inp)] if inp!=[] else []
not_partition(inp, 'is'), not_partition(inp, 'test'), not_partition(inp, 'what?')
没有任何功能。只需使用==
如果不允许使用任何内置函数，那么其中一个选项就是进行字符串比较。我假设您也不能使用len（）内置函数
mystring = "This is python class and it is a unicorn"
keyword = 'is'
list_of_words = []
word = ''
key_len = str_len = 0
#to get length of keyword as you cannot use len(keyword)
for _ in keyword: key_len +=1
for _ in mystring: str_len +=1
i = 0
start_pos = 0
if keyword in mystring:
while i < str_len:
if mystring[i:i+key_len] == keyword:
if i != 0:
list_of_words.append(mystring[start_pos:i])
list_of_words.append(keyword)
i += key_len
start_pos = i
else:
i+=1
if start_pos != str_len and start_pos != 0:
list_of_words.append(mystring[start_pos:])
if list_of_words:
print (list_of_words)
else:
print (keyword, 'is not found in', mystring)
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
输入字：
mystring = "This is python class and it is a unicorn"
keyword = ' is'
mystring = "This is python class and it is a unicorn"
keyword = ' unicorn'
mystring = "This is python class and it is a unicorn"
keyword = ' as'
mystring = "This is python class and it is a unicorn"
keyword = 'This'
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
输入字：
mystring = "This is python class and it is a unicorn"
keyword = ' is'
mystring = "This is python class and it is a unicorn"
keyword = ' unicorn'
mystring = "This is python class and it is a unicorn"
keyword = ' as'
mystring = "This is python class and it is a unicorn"
keyword = 'This'
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
输入字：
mystring = "This is python class and it is a unicorn"
keyword = ' is'
mystring = "This is python class and it is a unicorn"
keyword = ' unicorn'
mystring = "This is python class and it is a unicorn"
keyword = ' as'
mystring = "This is python class and it is a unicorn"
keyword = 'This'
输出：
['This', ' is', ' python class']
['This', 'is', 'python class and it', 'is', 'a unicorn']
['This', ' is', ' python class and it', ' is', ' a unicorn']
['This is python class and it is a', ' unicorn']
as is not found in This is python class and it is a unicorn
['This', ' is python class and it is a unicorn']
使用查找功能
您可以按如下方式搜索关键字并创建列表：
mystring = mystring.split(keyword)
mystring.insert(1, keyword)
print(mystring)
mystring = "This is python class"
keyword = ' as'
l = []
x = mystring.find(keyword)
if x != -1:
l.append(mystring[:x])
l.append(keyword)
l.append(mystring[x+len(keyword):])
print (l)
else:
print (keyword, 'is not found in', mystring)
您也可以尝试使用正则表达式解决此问题
import re
keyword = 'is'
str1 = "This hello is to test"
partition = re.split(r'\s({0})\s'.format(keyword), str1)
partition
['This hello', 'is', 'to test']
找到单词在列表中的位置。然后将索引前的所有值分配给before_关键字，将搜索词的index+len后的所有值分配给after_关键字您是说不能使用find
作为选项吗？如果找不到关键字，它将提供不正确的结果。此外，关键字可能并不总是IST中的第二个元素，这是一个庞大的代码：P@KuldeepSinghSidhu哈哈，是的：-）为什么第10行string=''
是doing@Nons我一直在向字符串
变量添加值，当它看到一个单词与关键字
变量相同时（在本例中，is
），它将把string
变量作为一个元素添加到列表中，然后添加关键字（is
），然后将string
变量重置为空字符串。提问者要求的解决方案不需要任何python的“内置方法”（如果我理解正确的话）.strip
，split
等您在解决方案中使用的方法是“内置方法”…是的。没有内置函数可以使用。因此我们的解决方案必须没有任何这些选项。是的！如果你想让我也写这些，请告诉我！not_strip
和not_split
：P但是，你也必须做not_len
可能not_append
~），还必须使用拆分字符串：
。。。它将成为一个全新的世界！如果OP要求，我将更新！我们都上当了。我们不能使用内置函数。这就像是治疗。编写代码时不使用任何内置函数：）“这是python类，它是独角兽”不起作用。多次检查关键字。嗯，我没想过。抢手货我一找到关键字就要跳出循环。修复了一个关键字多次出现的问题。还检查单词是字符串中的第一个还是最后一个单词。如果单词是第一个，则忽略上一个。若单词是最后一个，那个么after将被忽略。如果要打印空字符串以表示开头或后面，请告诉我。请注意，像as
和is
这样的单词将被拆分为类
和此
。如果要消除这种情况，代码可能会变得更复杂。我知道这是一个python类练习，因此与您一起研究解决方案。在现实世界中，您将使用内置的方法/函数来实现这一点。