I am having a number of strings in the loop, some strings contain "," and some doesn't contain, I want my code to check if there is any "," present in a string remove them and then print the string, and if its not present, print the string as it is.
here is my code:
for x in range(y):
c = containers[x].find("div",{"class":"cong-data"})
meeting = c.p.next_element
print(meeting)
Thanks in advance.
I recommend using Regular Expression sub() function.
import re
string = 'aabcdefg,hijklmnop,qrstu,ssads'
remove_commas = re.sub(',', '', string)
print(string)
print(remove_commas)
output:
aabcdefg,hijklmnop,qrstu,ssads
aabcdefghijklmnopqrstussads
Related
I want to tokenize string up to 3rd occurence of some delimiter and then return the rest of the string as last element of the tokenize array.
Example:
I have a String which looks like this:
String someString= 1.22.33.4
Now im tokenizing it by delimiter '.' like this:
def (a, b, c, d) = someString.tokenize('.')
And it works, but only if number of dots are exactly 3.
Now if someone puts more number of dots like:
String someString = 1.22.33.4.55
Then it wouldn't work, because the number of variables won't match. So i want to make sure it only tokenizes up to 3rd dot, and then gives back whatever is left. So what i want to achieve in this case would be:
a = 1, b=22, c=33, d=4.55
How to do that?
You can use the version of split with the second argument to restrict
the returned items. E.g.
def (a,b,c,d) = '1.22.33.4.55'.split("\\.", 4)
assert ["1","22","33","4.55"] == [a,b,c,d]
Not a one liner but it works:
String someString= '1.22.33.4.55'
def stringArray = someString.tokenize('.')
def (a,b,c) = stringArray
def d = stringArray.drop(3).join('.')
println "a=$a, b=$b, c=$c, d=$d"
result:
a=1, b=22, c=33, d=4.55
I would like to know how to replace non-numeric characters in a single string with different random integers.
I have tried the following:
text = '1$1#387'
rec_1 = re.sub("\D+",str(random.randint(0,9)),text)
It then produced:
output: 1717387
As you can see, the non-numeric characters have been replaced by the same integer. I would like each non-numeric character to be replaced by a different integer. For example:
desired output: 1714387
Please assist.
Use a function as the replacement value:
def replacement(match):
return str(random.randint(0, 9))
text = '1$1#387'
rec_1 = re.sub(r"\D", replacement, text)
rec_1 is now "1011387", or "1511387", ...
That's because the randint function is called only 1 time.
You can use a lambda to get a new randint each time:
rec_1 = re.sub("\D+", lambda x: str(random.randint(0, 9)), text)
My question is more or less similar to:
Is there a way to substring a string in Python?
but it's more specifically oriented.
How can I get a par of a string which is located between two known words in the initial string.
Example:
mySrting = "this is the initial string"
Substring = "initial"
knowing that "the" and "string" are the two known words in the string that can be used to get the substring.
Thank you!
You can start with simple string manipulation here. str.index is your best friend there, as it will tell you the position of a substring within a string; and you can also start searching somewhere later in the string:
>>> myString = "this is the initial string"
>>> myString.index('the')
8
>>> myString.index('string', 8)
20
Looking at the slice [8:20], we already get close to what we want:
>>> myString[8:20]
'the initial '
Of course, since we found the beginning position of 'the', we need to account for its length. And finally, we might want to strip whitespace:
>>> myString[8 + 3:20]
' initial '
>>> myString[8 + 3:20].strip()
'initial'
Combined, you would do this:
startIndex = myString.index('the')
substring = myString[startIndex + 3 : myString.index('string', startIndex)].strip()
If you want to look for matches multiple times, then you just need to repeat doing this while looking only at the rest of the string. Since str.index will only ever find the first match, you can use this to scan the string very efficiently:
searchString = 'this is the initial string but I added the relevant string pair a few more times into the search string.'
startWord = 'the'
endWord = 'string'
results = []
index = 0
while True:
try:
startIndex = searchString.index(startWord, index)
endIndex = searchString.index(endWord, startIndex)
results.append(searchString[startIndex + len(startWord):endIndex].strip())
# move the index to the end
index = endIndex + len(endWord)
except ValueError:
# str.index raises a ValueError if there is no match; in that
# case we know that we’re done looking at the string, so we can
# break out of the loop
break
print(results)
# ['initial', 'relevant', 'search']
You can also try something like this:
mystring = "this is the initial string"
mystring = mystring.strip().split(" ")
for i in range(1,len(mystring)-1):
if(mystring[i-1] == "the" and mystring[i+1] == "string"):
print(mystring[i])
I suggest using a combination of list, split and join methods.
This should help if you are looking for more than 1 word in the substring.
Turn the string into array:
words = list(string.split())
Get the index of your opening and closing markers then return the substring:
open = words.index('the')
close = words.index('string')
substring = ''.join(words[open+1:close])
You may want to improve a bit with the checking for the validity before proceeding.
If your problem gets more complex, i.e multiple occurrences of the pair values, I suggest using regular expression.
import re
substring = ''.join(re.findall(r'the (.+?) string', string))
The re should store substrings separately if you view them in list.
I am using the spaces between the description to rule out the spaces between words, you can modify to your needs as well.
Hello I am fairly new at programming and python and I have a question.
How would I go about printing or returning only numbers from a string
For example:
"Hu765adjH665Sdjda"
output:
"765665"
You can use re.sub to remove any character that is not a number.
import re
string = "Hu765adjH665Sdjda"
string = re.sub('[^0-9]', '', string)
print string
#'765665'
re.sub scan the string from left to right. everytime it finds a character that is not a number it replaces it for the empty string (which is the same as removing it for all practical purpose).
>>> s = "Hu765adjH665Sdjda"
>>> ''.join(c for c in s if c in '0123456789')
'765665'
a='a345g5'
for i in a:
if int(i.isnumeric()):
print(i,end=' ')
Try filter
>>> str='1qaz2wsx3edc4rfv5tgb6yhn7ujm8ik9ol'
>>> print str
1qaz2wsx3edc4rfv5tgb6yhn7ujm8ik9ol
>>> filter(lambda x:x>='0' and x<='9', str)
'123456789'
sentence = "Hu765adjH665Sdjda"
for number in sentence:
if number in "0123456789":
print(number)
How do trim off characters in a string, by how much you want?
For example, say your string is "Tony", but you wanted to display "ny" by trimming of the first two characters, how can this be done?
Sub Main()
Dim s As String
Dim Result As String
s = "Tony"
Result = LTrim(s)
msgbox(Result)
I have this so far using the LTrim function, so how do you specify by how much you want to cut to just display "ny" in the MessageBox?
You don't want LTrim. You want Right:
Result = Right(s, Len(s) - 2);
This will take all but the two left-most characters of s.
You could use the additional string functions to do the same thing,
for example:
X$ = RIGHT$(V$, 2) ' get the ending 2 chars of string
X$ = LEFT$(V$, 2) ' get the leading 2 chars of string
X$ = MID$(V$, 2, 2) ' get 2 chars from the inside of string
Well... If I was trying to clip off the beginning of a string, I would use two functions: StrReverse, and Remove.
I would first reverse the string, then use the remove function to cut off what is now the end, Then flip the remaining string back to it's original state using the reverse function again.
The code would look something like this:
Dim s As String = "Anthony"
Dim index As Integer = 2
Debug.Print(StrReverse(StrReverse(s).Remove(2)))
The output of this would be "ny" and the length will correspond to the index.