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How can I check a string for two letters or more?

I am pulling data from a table that changes often using Python - and the method I am using is not ideal. What I would like to have is a method to pull all strings that contain only one letter and leave out anything that is 2 or more.
An example of data I might get:
115
19A6
HYS8
568
In this example, I would like to pull 115, 19A6, and 568.
Currently I am using the isdigit() method to determine if it is a digit and this filters out all numbers with one letter, which works for some purposes, but is less than ideal.

Try this:
string_list = ["115", "19A6", "HYS8", "568"]
output_list = []
for item in string_list: # goes through the string list
letter_counter = 0
for letter in item: # goes through the letters of one string
if not letter.isdigit(): # checks if the letter is a digt
letter_counter += 1
if letter_counter < 2: # if the string has more then 1 letter it wont be in output list
output_list.append(item)
print(output_list)
Output:
['115', '19A6', '568']

Here is a one-liner with a regular expression:
import re
data = ["115", "19A6", "HYS8", "568"]
out = [string for string in data if len(re.sub("\d", "", string))<2]
print(out)
Output:
['115', '19A6', '568']

This is an excellent case for regular expressions (regex), which is available as the built-in re library.
The code below follows the logic:
Define the dataset. Two examples have been added to show that a string containing two alpha-characters is rejected.
Compile a character pattern to be matched. In this case, zero or more digits, followed by zero or one upper case letter, ending with zero of more digits.
Use the filter function to detect matches in the data list and output as a list.
For example:
import re
data = ['115', '19A6', 'HYS8', '568', 'H', 'HI']
rexp = re.compile('^\d*[A-Z]{0,1}\d*$')
result = list(filter(rexp.match, data))
print(result)
Output:
['115', '19A6', '568', 'H']

Another solution, without re using str.maketrans/str.translate:
lst = ["115", "19A6", "HYS8", "568"]
d = str.maketrans(dict.fromkeys(map(str, range(10)), ""))
out = [i for i in lst if len(i.translate(d)) < 2]
print(out)
Prints:
['115', '19A6', '568']

z=False
a = str(a)
for I in range(len(a)):
if a[I].isdigit():
z = True
break
else:
z="no digit"
print(z)```

How to remove unwanted elements from list using python

I have an input list:
list_1 = ['29','560001','08067739333','560037002','29AAACC0462F1Z0','55XX1XXX19','07S23X09','98561XXX1X9']
I have tried:
output_list = [i for i in list_1 if 'X' in i or i.isnumeric()==True]
Giving out with extra element '07S23X09' which is wrong:
output_list = ['29','560001','08067739333','560037002','55XX1XXX19','07S23X09','98561XXX1X9']
Expected output is the list with numbers and the elements with numbers and specific character X, else other elements should be discarded:
output_list = ['29','560001','08067739333','560037002','55XX1XXX19','98561XXX1X9']

You may use
import re
list_1 = ['29','560001','08067739333','560037002','29AAACC0462F1Z0','55XX1XXX19','07S23X09','98561XXX1X9']
rx = re.compile('[0-9X]+')
print ( [x for x in list_1 if rx.fullmatch(x)] )
# => ['29', '560001', '08067739333', '560037002', '55XX1XXX19', '98561XXX1X9']
See the Python demo.
With re.fullmatch('[0-9X]+', x), you only keep the items that fully consist of digits or X chars.
See ^[0-9X]+$ the regex demo.
NOTE: If there must be at least one digit in the string, i.e. if you want to fail and thus discard all items that are just XXX, you may use
^X*[0-9][0-9X]*$
Or, ^(?=X*[0-9])[0-9X]+$. See the regex demo.

How about:
output_list = [i for i in list_1 if i.replace('X', '').isnumeric()==True]
You seem to want all the numeric ones, but are ok if it's an 'X' in there. So if you remove the X's and check for numeric, that would do the trick.

This could be easily done with python regex like the following
import re
list_1 = ['29','560001','08067739333','560037002','29AAACC0462F1Z0','55XX1XXX19','07S23X09','98561XXX1X9']
l2 = re.sub('[a-zA-VY-Z]', "", str(list_1)) # delete unwanted characters which are lowercase and uppercase letters from a to v and Y-Z since we only need to preserve uppercase letter X
print(l2)
output
['29', '560001', '08067739333', '560037002', '29046210', '55XX1XXX19', '0723X09', '98561XXX1X9']

Replace a substring in a list of str only when the substring is alone - python

Let us suppose I have a list of string:
l = ['firm incorp', 'firm2 inc']
and I only want to remove the suffix 'inc' from the second entry but if I do the following
l_r = [s.replace('inc', '') for s in l]
I'll remove the substring inc from the first entry as well. What is the most effective way to handle this?
NB: the suffix I want to remove are, in principle, only at the end of the string.

Remove the string only found at word boundaries using regular expression:
import re
l_r = [re.sub(r'\binc$', '', s) for s in l]
\b is special character matching word boundaries, $ means the end of the string or just before the newline at the end of the string. For further information read the re docs.

Here is a solution without regex:
l = ['firm incorp', 'firm2 inc']
l2 = [' '.join(j for j in i.split(' ') if j != 'inc') for i in l]
# ['firm incorp', 'firm2']

>>> items = ['firm incorp', 'firm2 inc']
>>> [s[:-4] if s.endswith(' inc') else s for s in items]
['firm incorp', 'firm2']

Python find element in list that ends with number

I have a list of strings, and I want to all the strings that end with _1234 where 1234 can be any 4-digit number. It's ideal to find all the elements, and what the digits actually are, or at least return the 1st matching element, and what the 4 digit is.
For example, I have
['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765']
I want to get
['1024', '0510']
Okay so far I got, _\d{4}$ will match _1234 and return a match object, and the match_object.group(0) is the actual matched string. But is there a better way to look for _\d{4}$ but only return \d{4} without the _?

Use re.search():
import re
lst = ['A', 'BB_1024', 'CQ_2', 'x_0510']
newlst = []
for item in lst:
match = re.search(r'_(\d{4})\Z', item)
if match:
newlst.append(match.group(1))
print(newlst) # ['1024', '0510']
As for the regex, the pattern matches an underscore and exactly 4 digits at the end of the string, capturing only the digits (note the parens). The captured group is then accessible via match.group(1) (remember that group(0) is the entire match).

import re
src = ['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765', 'AB2421', 'D3&1345']
res = []
p = re.compile('.*\D(\d{4})$')
for s in src:
m = p.match(s)
if m:
res.append(m.group(1))
print(res)
Works fine, \D means not a number, so it will match 'AB2421', 'D3&1345' and so on.

Please show some code next time you ask a question here, even if it doesn't work at all. It makes it easier for people to help you.
If you're interested in a solution without any regex, here's a way with list comprehensions:
>>> data = ['A', 'BB_1024', 'CQ_2', 'x_0510', 'y_98765']
>>> endings = [text.split('_')[-1] for text in data]
>>> endings
['A', '1024', '2', '0510', '98765']
>>> [x for x in endings if x.isdigit() and len(x)==4]
['1024', '0510']

Try this:
[s[-4:] for s in lst if s[-4:].isdigit() and len(s) > 4]
Just check the last four characters if it's a number or not.
added the len(s) > 4 to correct the mistake Joran pointed out.

Try this code:
r = re.compile(".*?([0-9]+)$")
newlist = filter(r.match, mylist)
print newlist

Filter strings where there are n equal characters in a row

Is there an option how to filter those strings from list of strings which contains for example 3 equal characters in a row? I created a method which can do that but I'm curious whether is there a more pythonic way or more efficient or more simple way to do that.
list_of_strings = []
def check_3_in_row(string):
for ch in set(string):
if ch*3 in string:
return True
return False
new_list = [x for x in list_of_strings if check_3_in_row(x)]
EDIT:
I've just found out one solution:
new_list = [x for x in set(keywords) if any(ch*3 in x for ch in x)]
But I'm not sure which way is faster - regexp or this.

You can use Regular Expression, like this
>>> list_of_strings = ["aaa", "dasdas", "aaafff", "afff", "abbbc"]
>>> [x for x in list_of_strings if re.search(r'(.)\1{2}', x)]
['aaa', 'aaafff', 'afff', 'abbbc']
Here, . matches any character and it is captured in a group ((.)). And we check if the same captured character (we use the backreference \1 refer the first captured group in the string) appears two more times ({2} means two times).