I am trying to remove a specific word from within a sentence, which is 'you'. The code is as listed below:
out1.text_condition = out1.text_condition.replace('you','')
This works, however, it also removes it from within a word that contains it, so when 'your' appears, it removes the 'you' from within it, leaving 'r' standing. Can anyone help me figure out what I can do to just remove the word, not the letters from within another string?
Thanks!
In order to replace whole words and not substrings, you should use a regular expression (regex).
Here is how to replace a whole word with the module re:
import re
def replace_whole_word_from_string(word, string, replacement=""):
regular_expression = rf"\b{word}\b"
return re.sub(regular_expression, replacement, string)
string = "you you ,you your"
result = replace_whole_word_from_string("you", string)
print(result)
Output:
, your
Explanation:
The two \b are what we call "word boundaries". The advantage over str.replace is that it will take into account the punctuation too.
In order to create the regular expression, here we use Literal String Interpolation (also called "f-strings", https://www.python.org/dev/peps/pep-0498/).
To create a "f-string", we add the prefix f.
We also use the prefix r, in order to create a "raw string". We use a raw string in order to avoid escaping the backslash in \b.
Without the prefix r, we would have written regular_expression = f"\\b{word}\\b".
If you had used string.replace(' you ', ' '), you would have received this (wrong) output:
you ,you your
A very simple solution is to replace the word with spaces around it with one space:
out1.text_condition = out1.text_condition.replace(' you ', ' ')
But note that it wouldn't remove for example you. (in the end of the sentence) or you,, etc.
Easiest way is probably just to assume there are spaces before and after the word:
out1.text_condition = out1.text_condition.replace(' you ','')
Related
I would like to know how to remove specific words from a string in python without deleting them from other words they composed.
For example if I want to remove 'is' from the following sentence:
s = 'isabelle is in Paris'
The .replace() function delete 'is' in 'isabelle' and in 'Paris':
s = 'isabelle is in Paris'
s.replace('is', '')
It gives me abelle in Par But I want isabelle in Paris. Is there a way to delete only 'is'?
I tried: s.replace(' is ', '') with a space each side of 'is' but in this case 'is' is not removed in the string s = 'Isabelle is, as you know, in Paris'
Thank you
Use a regular expression instead of replacing an ordinary string. You can then use \b in the regexp to match word boundaries.
import re
s = re.sub(r'\bis\b', '', s)
I have a long regex with many alternations and I want to be able to replace each match from the regex with itself followed by a new line ('\n').
What is the most efficient way to do so with re.sub()?
Here is a simple example:
s = 'I want to be able to replace many words, especially in this sentence, since it will help me solve by problem. That makes sense right?'
pattern = re.compile(r'words[,]|sentence[,]|problem[.]')
for match in matches:
re.sub(pattern, match + '\n', match)
I know this for loop will not work, I am just hoping to clarify what I am trying to solve here. Thanks in advance for any help. I may be missing something very straightforward.
To replace a whole match with itself you may use a replacement backreference \g<0>. However, you want to replace and store the matches inside a variable. You need to pass a callback method as a replacement argument to re.sub, and return the whole match value (match.group()) with a newline appended to the value:
import re
matches = [] # Variable to hold the matches
def repl(m): # m is a match data object
matches.append(m.group()) # Add a whole match value
return "{}\n".format(m.group()) # Return the match and a newline appended to it
s = 'I want to be able to replace many words, especially in this sentence, since it will help me solve by problem. That makes sense right?'
pattern = re.compile(r'words[,]|sentence[,]|problem[.]')
s = re.sub(pattern, repl, s)
print(s)
print(matches)
See the Python demo
Just like this?
text ='I want to be able to replace many words, especially in this sentence, since it will help me solve by problem. That makes sense right?'
text_list = tex t.replace('.',',').strip(',|.|?').split(',')
##Remove the beginning and end symbols.And split by ','
print (text_list)
for i in text_list:
ii=i.split(',')
print(ii)
Result
['I want to be able to replace many words', ' especially in this sentence', ' since it will help me solve by problem', ' That makes sense right']
['I want to be able to replace many words']
[' especially in this sentence']
[' since it will help me solve by problem']
[' That makes sense right']
the second parameter of re.sub can either be a string or a callable that takes in the match instance and returns a string. so do this
def break_line(match):
return "\n" + match.group()
text = re.sub(pattern, break_line, text)
for string "//div[#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings']", I want to find "#..'...'" like "#id~'objectnavigator-card-list'" or "#class~'outbound-alert-settings'". But when I use regex ((#.+)\~(\'.*?\')), it find "#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings'". So how to modify the regex to find the string successfully?
Use non-capturing, non greedy, modifiers on the inner brackets and search for not the terminating character, e.g.:
re.findall(r"((?:#[^\~]+)\~(?:\'[^\]]*?\'))", test)
On your test string returns:
["#id~'objectnavigator-card-list'", "#class~'outbound-alert-settings'"]
Limit the characters you want to match between the quotes to not match the quote:
>>> re.findall(r'#[a-z]+~\'[-a-z]*\'', x)
I find it's much easier to look for only the characters I know are going to be in a matching section rather than omitting characters from more permissive matches.
For your current test string's input you can try this pattern:
import re
a = "//div[#id~'objectnavigator-card-list']//li[#class~'outbound-alert-settings']"
# find everything which begins by '#' and neglect ']'
regex = re.compile(r'(#[^\]]+)')
strings = re.findall(regex, a)
# Or simply:
# strings = re.findall('(#[^\\]]+)', a)
print(strings)
Output:
["#id~'objectnavigator-card-list'", "#class~'outbound-alert-settings'"]
I am trying to do the following with a regular expression:
import re
x = re.compile('[^(going)|^(you)]') # words to replace
s = 'I am going home now, thank you.' # string to modify
print re.sub(x, '_', s)
The result I get is:
'_____going__o___no______n__you_'
The result I want is:
'_____going_________________you_'
Since the ^ can only be used inside brackets [], this result makes sense, but I'm not sure how else to go about it.
I even tried '([^g][^o][^i][^n][^g])|([^y][^o][^u])' but it yields '_g_h___y_'.
Not quite as easy as it first appears, since there is no "not" in REs except ^ inside [ ] which only matches one character (as you found). Here is my solution:
import re
def subit(m):
stuff, word = m.groups()
return ("_" * len(stuff)) + word
s = 'I am going home now, thank you.' # string to modify
print re.sub(r'(.+?)(going|you|$)', subit, s)
Gives:
_____going_________________you_
To explain. The RE itself (I always use raw strings) matches one or more of any character (.+) but is non-greedy (?). This is captured in the first parentheses group (the brackets). That is followed by either "going" or "you" or the end-of-line ($).
subit is a function (you can call it anything within reason) which is called for each substitution. A match object is passed, from which we can retrieve the captured groups. The first group we just need the length of, since we are replacing each character with an underscore. The returned string is substituted for that matching the pattern.
Here is a one regex approach:
>>> re.sub(r'(?!going|you)\b([\S\s]+?)(\b|$)', lambda x: (x.end() - x.start())*'_', s)
'_____going_________________you_'
The idea is that when you are dealing with words and you want to exclude them or etc. you need to remember that most of the regex engines (most of them use traditional NFA) analyze the strings by characters. And here since you want to exclude two word and want to use a negative lookahead you need to define the allowed strings as words (using word boundary) and since in sub it replaces the matched patterns with it's replace string you can't just pass the _ because in that case it will replace a part like I am with 3 underscore (I, ' ', 'am' ). So you can use a function to pass as the second argument of sub and multiply the _ with length of matched string to be replace.
How to match the following i want all the names with in the single quotes
This hasn't been much that much of a twist and turn's to 'Tom','Harry' and u know who..yes its 'rock'
How to extract the name within the single quotes only
name = re.compile(r'^\'+\w+\'')
The following regex finds all single words enclosed in quotes:
In [6]: re.findall(r"'(\w+)'", s)
Out[6]: ['Tom', 'Harry', 'rock']
Here:
the ' matches a single quote;
the \w+ matches one or more word characters;
the ' matches a single quote;
the parentheses form a capture group: they define the part of the match that gets returned by findall().
If you only wish to find words that start with a capital letter, the regex can be modified like so:
In [7]: re.findall(r"'([A-Z]\w*)'", s)
Out[7]: ['Tom', 'Harry']
I'd suggest
r = re.compile(r"\B'\w+'\B")
apos = r.findall("This hasn't been much that much of a twist and turn's to 'Tom','Harry' and u know who..yes its 'rock'")
Result:
>>> apos
["'Tom'", "'Harry'", "'rock'"]
The "negative word boundaries" (\B) prevent matches like the 'n' in words like Rock'n'Roll.
Explanation:
\B # make sure that we're not at a word boundary
' # match a quote
\w+ # match one or more alphanumeric characters
' # match a quote
\B # make sure that we're not at a word boundary
^ ('hat' or 'caret', among other names) in regex means "start of the string" (or, given particular options, "start of a line"), which you don't care about. Omitting it makes your regex work fine:
>>> re.findall(r'\'+\w+\'', s)
["'Tom'", "'Harry'", "'rock'"]
The regexes others have suggested might be better for what you're trying to achieve, this is the minimal change to fix your problem.
Your regex can only match a pattern following the start of the string. Try something like: r"'([^']*)'"