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Getting word from string

How can i get word example from such string:
str = "http://test-example:123/wd/hub"
I write something like that
print(str[10:str.rfind(':')])
but it doesn't work right, if string will be like
"http://tests-example:123/wd/hub"

You can use this regex to capture the value preceded by - and followed by : using lookarounds
(?<=-).+(?=:)
Regex Demo
Python code,
import re
str = "http://test-example:123/wd/hub"
print(re.search(r'(?<=-).+(?=:)', str).group())
Outputs,
example
Non-regex way to get the same is using these two splits,
str = "http://test-example:123/wd/hub"
print(str.split(':')[1].split('-')[1])
Prints,
example

You can use following non-regex because you know example is a 7 letter word:
s.split('-')[1][:7]
For any arbitrary word, that would change to:
s.split('-')[1].split(':')[0]

many ways
using splitting:
example_str = str.split('-')[-1].split(':')[0]
This is fragile, and could break if there are more hyphens or colons in the string.
using regex:
import re
pattern = re.compile(r'-(.*):')
example_str = pattern.search(str).group(1)
This still expects a particular format, but is more easily adaptable (if you know how to write regexes).

I am not sure why do you want to get a particular word from a string. I guess you wanted to see if this word is available in given string.
if that is the case, below code can be used.
import re
str1 = "http://tests-example:123/wd/hub"
matched = re.findall('example',str1)

Split on the -, and then on :
s = "http://test-example:123/wd/hub"
print(s.split('-')[1].split(':')[0])
#example

using re
import re
text = "http://test-example:123/wd/hub"
m = re.search('(?<=-).+(?=:)', text)
if m:
print(m.group())

Python strings has built-in function find:
a="http://test-example:123/wd/hub"
b="http://test-exaaaample:123/wd/hub"
print(a.find('example'))
print(b.find('example'))
will return:
12
-1
It is the index of found substring. If it equals to -1, the substring is not found in string. You can also use in keyword:
'example' in 'http://test-example:123/wd/hub'
True

Change part of a word (string) into in a different string if a sign occurs. Python

How most effectively do I cut out a part of a word if the character '=#=' appears and then finish cutting the word if the character '=#=' appears? For example:
From a large string
'321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
The python code returns:
'I-LOVE-STACK-OVER-FLOW'
Any help will be appreciated.

Using split():
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
st = '=#='
ed = '=#='
print((s.split(st))[1].split(ed)[0])
Using regex:
import re
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319'
print(re.search('%s(.*)%s' % (st, ed), s).group(1))
OUTPUT:
I-LOVE-STACK-OVER-FLOW

In addition to #DirtyBit's answer, if you want to also handle cases of more than 2 '=#='s, you can split the string, and then add every other element:
s = '321#5=85#45#41=#=I-LOVE-STACK-OVER-FLOW=#=3234#41#=q#$^1=#=xx$q=#=xpa$=4319=#=|I-ALSO-LOVE-SO=#=3123123'
parts = s.split('=#=')
print(''.join([parts[i] for i in range(1,len(parts),2)]))
Output
I-LOVE-STACK-OVER-FLOW|I-ALSO-LOVE-SO

The explanation is in the code.
import re
ori_list = re.split("=#=",ori_str)
# you can imagine your goal is to find the string wrapped between signs of "=#="
# so after the split, the even number position must be the parts outsides of "=#="
# and the odd number position is what you want
for i in range(len(ori_list)):
if i%2 == 1:#odd position
print(ori_list[i])

Parsing String by regular expression in python

How can I parse this string in python?
Input String:
someplace 2018:6:18:0 25.0114 95.2818 2.71164 66.8962 Entire grid contents are set to missing data
to this
Output array:
['someplace','2018:6:18:0','25.0114','95.2818','2.71164','66.8962','Entire grid contents are set to missing data']
I have already tried with split(' ') but as it is not clear how many spaces are between the sub-strings and inside the last sub-string there may be spaces so this doesn't work.
I need the regular expression.

If you do not provide a sep-character, pythons split(sep=None, maxsplit=-1) (doku) will treat consecutive whitespaces as one whitespace and split by those. You can limit the amount of splits to be done by providing a maxsplit value:
data = "someplace 2018:6:18:0 25.0114 95.2818 2.71164 66.8962 Entire grid contents are set to missing data"
spl = data.split(None,6) # dont give a split-char, use 6 splits at most
print(spl)
Output:
['someplace', '2018:6:18:0', '25.0114', '95.2818', '2.71164',
'66.8962', 'Entire grid contents are set to missing data']
This will work as long as the first text does not contain any whitespaces.
If the fist text may contain whitespaces, you can use/refine this regex solution:
import re
reg = re.findall(r"([^\d]+?) +?([\d:]+) +?([\d.]+) +?([\d.]+) +?([\d.]+) +?([\d.]+) +(.*)$",data)[0]
print(reg)
Output:
('someplace', '2018:6:18:0', '25.0114', '95.2818', '2.71164', '66.8962', 'Entire grid contents are set to missing data')
Use f.e.https://regex101.com to check/proof the regex against your other data (follow the link, it uses above regex on sample data)

[A-Z]{1}[a-zA-Z ]{15,45}|[\w|:|.]+
You can test it here https://pythex.org/
Modify 15,45 according to your needs.

Maxsplit works with re.split(), too:
import re
re.split(r"\s+",text,maxsplit=6)
Out:
['someplace',
'2018:6:18:0',
'25.0114',
'95.2818',
'2.71164',
'66.8962',
'Entire grid contents are set to missing data']
EDIT:
If the first and last text parts don't contain digits, we don't need maxsplit and do not have to rely on number of parts with consecutive spaces:
re.split("\s+(?=\d)|(?<=\d)\s+",s)
We cut the string where a space is followed by a digit or vice versa using lookahead and lookbehind.

It is hard to answer your question as the requirements are not very precise. I think I would split the line with the split() function and then join the items when their contents has no numbers. Here is a snippet that works with your lonely sample:
def containsNumbers(s):
return any(c.isdigit() for c in s)
data = "someplace 2018:6:18:0 25.0114 95.2818 2.71164 66.8962 Entire grid contents are set to missing data"
lst = data.split()
lst2 = []
i = 0
agg = ''
while i < len(lst):
if containsNumbers(lst[i]):
if agg != '':
lst2.append(agg)
agg = ''
lst2.append(lst[i])
else:
agg += ' '+lst[i]
agg = agg.strip()
if i == len(lst) - 1:
lst2.append(agg)
i += 1
print(lst2)

How to convert a multiline string into a list of lines?

In sikuli I've get a multiline string from clipboard like this...
Names = App.getClipboard();
So Name =
#corazona
#Pebleo00
#cofriasd
«paflio
and I have use this regex to delete the first character if it is not in x00-x7f hex range or is not a word, or is a digit
import re
Names = re.sub(r"(?m)^([^\x00-\x7F]+|\W|\d)", "", Names)
So now Names =
corazona
Pebleo00
cofriasd
paflio
But, I am having trouble with the second regex that converts "Names" into the items of a sequence. I would like to convert "Names" into...
'corazona', 'Pebleo00', 'cofriasd', 'paflio'
or
'corazona', 'Pebleo00', 'cofriasd', 'paflio',
So sikuli can then recognize it as a List (I've found that Sikuli is able to recognize it even with those last "comma" and "space" in the end) by using...
NamesAsList = eval(Names)
How could I do this in python? is it necessary to use regex, or there is other way to do this in python?
I have already done this but using .Net regex, I just don't know how to do it in python, I have googled it with no result.
This is how I did it using .Net regex
Text to find:
(.*[^$])(\r\n|\z)
Replace with:
'$1',%" "%
Thanks Advanced.

A couple of one liners. Your question isn't completely clear - but I am assuming - you want to split a given string delimited by 'newline' and then generate a list of strings by removing the first character if it's not alpha numeric. Here's how I'd go about it
import re
r = re.compile(r'^[a-zA-Z0-9]') # match # beginning anything that's not alpha numeric
s = '#abc\ndef\nghi'
l = [r.sub('', x) for x in s.split()]
# join this list with comma (if that's required else you got the list already)
','.join(l)
Hope that's what you want.

If Names is a string before you "convert" it, in which each name is separated by a new line ('\n'), then this will work:
NamesAsList = '\n'.split(Names)
See this question for other options.

You could use splitlines()
import re
clipBoard = App.getClipboard();
Names = re.sub(r"(?m)^([^\x00-\x7F]+|\W|\d)", "", clipBoard)
# Replace the end of a line with a comma.
singleNames = ', '.join(Names.splitlines())
print(singleNames)

python regex for repeating string

I am wanting to verify and then parse this string (in quotes):
string = "start: c12354, c3456, 34526; other stuff that I don't care about"
//Note that some codes begin with 'c'
I would like to verify that the string starts with 'start:' and ends with ';'
Afterward, I would like to have a regex parse out the strings. I tried the following python re code:
regx = r"start: (c?[0-9]+,?)+;"
reg = re.compile(regx)
matched = reg.search(string)
print ' matched.groups()', matched.groups()
I have tried different variations but I can either get the first or the last code but not a list of all three.
Or should I abandon using a regex?
EDIT: updated to reflect part of the problem space I neglected and fixed string difference.
Thanks for all the suggestions - in such a short time.

In Python, this isn’t possible with a single regular expression: each capture of a group overrides the last capture of that same group (in .NET, this would actually be possible since the engine distinguishes between captures and groups).
Your easiest solution is to first extract the part between start: and ; and then using a regular expression to return all matches, not just a single match, using re.findall('c?[0-9]+', text).

You could use the standard string tools, which are pretty much always more readable.
s = "start: c12354, c3456, 34526;"
s.startswith("start:") # returns a boolean if it starts with this string
s.endswith(";") # returns a boolean if it ends with this string
s[6:-1].split(', ') # will give you a list of tokens separated by the string ", "

This can be done (pretty elegantly) with a tool like Pyparsing:
from pyparsing import Group, Literal, Optional, Word
import string
code = Group(Optional(Literal("c"), default='') + Word(string.digits) + Optional(Literal(","), default=''))
parser = Literal("start:") + OneOrMore(code) + Literal(";")
# Read lines from file:
with open('lines.txt', 'r') as f:
for line in f:
try:
result = parser.parseString(line)
codes = [c[1] for c in result[1:-1]]
# Do something with teh codez...
except ParseException exc:
# Oh noes: string doesn't match!
continue
Cleaner than a regular expression, returns a list of codes (no need to string.split), and ignores any extra characters in the line, just like your example.

import re
sstr = re.compile(r'start:([^;]*);')
slst = re.compile(r'(?:c?)(\d+)')
mystr = "start: c12354, c3456, 34526; other stuff that I don't care about"
match = re.match(sstr, mystr)
if match:
res = re.findall(slst, match.group(0))
results in
['12354', '3456', '34526']