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Find the word from the list given and replace the words so found

My question is pretty simple, but I haven't been able to find a proper solution.
Given below is my program:
given_list = ["Terms","I","want","to","remove","from","input_string"]
input_string = input("Enter String:")
if any(x in input_string for x in given_list):
#Find the detected word
#Not in bool format
a = input_string.replace(detected_word,"")
print("Some Task",a)
Here, given_list contains the terms I want to exclude from the input_string.
Now, the problem I am facing is that the any() produces a bool result and I need the word detected by the any() and replace it with a blank, so as to perform some task.
Edit: any() function is not required at all, look for useful solutions below.

Iterate over given_list and replace them:
for i in given_list:
input_string = input_string.replace(i, "")
print("Some Task", input_string)

No need to detect at all:
for w in given_list:
input_string = input_string.replace(w, "")
str.replace will not do anything if the word is not there and the substring test needed for the detection has to scan the string anyway.

The problem with finding each word and replacing it is that python will have to iterate over the whole string, repeatedly. Another problem is you will find substrings where you don't want to. For example, "to" is in the exclude list, so you'd end up changing "tomato" to "ma"
It seems to me like you seem to want to replace whole words. Parsing is a whole new subject, but let's simplify. I'm just going to assume everything is lowercase with no punctuation, although that can be improved later. Let's use input_string.split() to iterate over whole words.
We want to replace some words with nothing, so let's just iterate over the input_string, and filter out the words we don't want, using the builtin function of the same name.
exclude_list = ["terms","i","want","to","remove","from","input_string"]
input_string = "one terms two i three want to remove"
keepers = filter(lambda w: w not in exclude_list, input_string.lower().split())
output_string = ' '.join(keepers)
print (output_string)
one two three
Note that we create an iterator that allows us to go through the whole input string just once. And instead of replacing words, we just basically skip the ones we don't want by having the iterator not return them.
Since filter requires a function for the boolean check on whether to include or exclude each word, we had to define one. I used "lambda" syntax to do that. You could just replace it with
def keep(word):
return word not in exclude_list
keepers = filter(keep, input_string.split())

To answer your question about any, use an assignment expression (Python 3.8+).
if any((word := x) in input_string for x in given_list):
# match captured in variable word

How do I find index in string without using find() in python

I'm currently using the find function and found a slight problem.
theres gonna be a fire here
If I have a sentence with the word "here" and "theres" and I use find() to find "here"s index, I instead get "theres"
I thought find() would be like
if thisword in thatword:
as it would find the word, not a substring within a string.
Is there another function that may work similarly? I'm using find() quite heavily would like to know of alternatives before I clog the code with string.split() then iterate until I find the exact match with an index counter on the side.
MainLine = str('theres gonna be a fire here')
WordtoFind = str('here')
#String_Len = MainLine.find(WordtoFind)
split_line = MainLine.split()
indexCounter = 0
for i in range (0,len(split_line)):
indexCounter += (len(split_line[i]) + 1)
if WordtoFind in split_line[i]:
#String_Len = MainLine.find(split_line[i])
String_Len = indexCounter
break

The best route would be regular expressions. To find a "word" just make sure that the leading and ending characters are not alphanumeric. It uses no splits, has no exposed loops, and even works when you run into a weird sentence like "There is a fire,here". A find_word function might look like this
import re
def find_word_start(word, string):
pattern = "(?<![a-zA-Z0-9])"+word+"(?![a-zA-Z0-9])"
result = re.search(pattern, string)
return result.start()
>> find_word_start("here", "There is a fire,here")
>> 16
The regex I made uses a trick called lookarounds that make sure that the characters preceding and after the word are not letters or numbers. https://www.regular-expressions.info/lookaround.html. The term [a-zA-Z0-9] is a character set that is comprised of a single character in the sets a-z, A-Z, and 0-9. Look up the python re module to find out more about regular expressions.

Is it possible to search and replace a string with "any" characters?

There are probably several ways to solve this problem, so I'm open to any ideas.
I have a file, within that file is the string "D133330593" Note: I do have the exact position within the file this string exists, but I don't know if that helps.
Following this string, there are 6 digits, I need to replace these 6 digits with 6 other digits.
This is what I have so far:
def editfile():
f = open(filein,'r')
filedata = f.read()
f.close()
#This is the line that needs help
newdata = filedata.replace( -TOREPLACE- ,-REPLACER-)
#Basically what I need is something that lets me say "D133330593******"
#->"D133330593123456" Note: The following 6 digits don't need to be
#anything specific, just different from the original 6
f = open(filein,'w')
f.write(newdata)
f.close()

Use the re module to define your pattern and then use the sub() function to substitute occurrence of that pattern with your own string.
import re
...
pat = re.compile(r"D133330593\d{6}")
re.sub(pat, "D133330593abcdef", filedata)
The above defines a pattern as -- your string ("D133330593") followed by six decimal digits. Then the next line replaces ALL occurrences of this pattern with your replacement string ("abcdef" in this case), if that is what you want.
If you want a unique replacement string for each occurrence of pattern, then you could use the count keyword argument in the sub() function, which allows you to specify the number of times the replacement must be done.
Check out this library for more info - https://docs.python.org/3.6/library/re.html

Let's simplify your problem to you having a string:
s = "zshisjD133330593090909fdjgsl"
and you wanting to replace the 6 characters after "D133330593" with "123456" to produce:
"zshisjD133330594123456fdjgsl"
To achieve this, we can first need to find the index of "D133330593". This is done by just using str.index:
i = s.index("D133330593")
Then replace the next 6 characters, but for this, we should first calculate the length of our string that we want to replace:
l = len("D133330593")
then do the replace:
s[:i+l] + "123456" + s[i+l+6:]
which gives us the desired result of:
'zshisjD133330593123456fdjgsl'
I am sure that you can now integrate this into your code to work with a file, but this is how you can do the heart of your problem .
Note that using variables as above is the right thing to do as it is the most efficient compared to calculating them on the go. Nevertheless, if your file isn't too long (i.e. efficiency isn't too much of a big deal) you can do the whole process outlined above in one line:
s[:s.index("D133330593")+len("D133330593")] + "123456" + s[s.index("D133330593")+len("D133330593")+6:]
which gives the same result.

How to use Boolean OR inside a regex

I want to use a regex to find a substring, followed by a variable number of characters, followed by any of several substrings.
an re.findall of
"ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
should give me:
['ATGTCAGGTAA', 'ATGTCAGGTAAGCTTAG', 'ATGTCAGGTAAGCTTAGGGCTTTAG']
I have tried all of the following without success:
import re
string2 = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
re.findall('(ATG.*TAA)|(ATG.*TAG)', string2)
re.findall('ATG.*(TAA|TAG)', string2)
re.findall('ATG.*((TAA)|(TAG))', string2)
re.findall('ATG.*(TAA)|(TAG)', string2)
re.findall('ATG.*(TAA)|ATG.*(TAG)', string2)
re.findall('(ATG.*)(TAA)|(ATG.*)(TAG)', string2)
re.findall('(ATG.*)TAA|(ATG.*)TAG', string2)
What am I missing here?

This is not super-easy, because a) you want overlapping matches, and b) you want greedy and non-greedy and everything inbetween.
As long as the strings are fairly short, you can check every substring:
import re
s = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
p = re.compile(r'ATG.*TA[GA]$')
for start in range(len(s)-6): # string is at least 6 letters long
for end in range(start+6, len(s)):
if p.match(s, pos=start, endpos=end):
print(s[start:end])
This prints:
ATGTCAGGTAA
ATGTCAGGTAAGCTTAG
ATGTCAGGTAAGCTTAGGGCTTTAG
Since you appear to work with DNA sequences or something like that, make sure to check out Biopython, too.

I like the accepted answer just fine :-) That is, I'm adding this for info, not looking for points.
If you have heavy need for this, trying a match on O(N^2) pairs of indices may soon become unbearably slow. One improvement is to use the .search() method to "leap" directly to the only starting indices that can possibly pay off. So the following does that.
It also uses the .fullmatch() method so that you don't have to artificially change the "natural" regexp (e.g., in your example, no need to add a trailing $ to the regexp - and, indeed, in the following code doing so would no longer work as intended). Note that .fullmatch() was added in Python 3.4, so this code also requires Python 3!
Finally, this intends to generalize the re module's finditer() function/method. While you don't need match objects (you just want strings), they're far more generally applicable, and returning a generator is often friendlier than returning a list too.
So, no, this doesn't do exactly what you want, but does things from which you can get what you want, in Python 3, faster:
def finditer_overlap(regexp, string):
start = 0
n = len(string)
while start <= n:
# don't know whether regexp will find shortest or
# longest match, but _will_ find leftmost match
m = regexp.search(string, start)
if m is None:
return
start = m.start()
for finish in range(start, n+1):
m = regexp.fullmatch(string, start, finish)
if m is not None:
yield m
start += 1
Then, e.g.,
import re
string2 = "ATGTCAGGTAAGCTTAGGGCTTTAGGATT"
pat = re.compile("ATG.*(TAA|TAG)")
for match in finditer_overlap(pat, string2):
print(match.group())
prints what you wanted in your example. The other ways you tried to write a regexp should also work. In this example it's faster because the second time around the outer loop start is 1, and regexp.search(string, 1) fails to find another match, so the generator exits at once (so skips checking O(N^2) other index pairs).

replace multiple words - python

There can be an input "some word".
I want to replace this input with "<strong>some</strong> <strong>word</strong>" in some other text which contains this input
I am trying with this code:
input = "some word".split()
pattern = re.compile('(%s)' % input, re.IGNORECASE)
result = pattern.sub(r'<strong>\1</strong>',text)
but it is failing and i know why: i am wondering how to pass all elements of list input to compile() so that (%s) can catch each of them.
appreciate any help

The right approach, since you're already splitting the list, is to surround each item of the list directly (never using a regex at all):
sterm = "some word".split()
result = " ".join("<strong>%s</strong>" % w for w in sterm)
In case you're wondering, the pattern you were looking for was:
pattern = re.compile('(%s)' % '|'.join(sterm), re.IGNORECASE)
This works on your string because the regular expression would become
(some|word)
which means "matches some or matches word".
However, this is not a good approach as it does not work for all strings. For example, consider cases where one word contains another, such as
a banana and an apple
which becomes:
<strong>a</strong> <strong>banana</strong> <strong>a</strong>nd <strong>a</strong>n <strong>a</strong>pple

It looks like you're wanting to search for multiple words - this word or that word. Which means you need to separate your searches by |, like the script below:
import re
text = "some word many other words"
input = '|'.join('some word'.split())
pattern = re.compile('(%s)' % input, flags=0)
print pattern.sub(r'<strong>\1</strong>',text)

I'm not completely sure if I know what you're asking but if you want to pass all the elements of input in as parameters in the compile function call, you can just use *input instead of input. * will split the list into its elements. As an alternative, could't you just try joining the list with and adding at the beginning and at the end?

Alternatively, you can use the join operator with a list comprehension to create the intended result.
text = "some word many other words".split()
result = ' '.join(['<strong>'+i+'</strong>' for i in text])