I have a series of strings, most of which contain 4 digits in a row. I want to slice the string at the end of that fourth digit, using Python. Sometimes the string contains more than one such pattern. What I want is the index position of the FIRST match of my regular expression. What I have been able to get is the LAST match.
myString = 'Today is June 14, 2019. I sometimes like to think back when I was a child in 1730.'
theYear = re.compile("\d{4}")
[(m.start(0), m.end(0)) for m in re.finditer(theYear, myString)]
print m.span(0)
The result is (77, 81), which is the index position for the second date, not the first one. I know the problem is my loop, which will iterate through all of the matches, leaving me with the last one. But I havn't been able to figure out how to access those index positions without looping.
Thanks for any help.
print theYear.search(myString).span()
Related
I have values 1000+ rows with variable values entered as below
5.99
5.188.0
v5.33
v.440.0
I am looking in Gsheet another column to perform following operations:
Remove the 'v' character from the values
if there is 2nd '.' missing as so string can become 5.88 --> 5.88.0
Can help please in the regex and replace logic as tried this but new to regex making. Thanks for the help given
=regexmatch(<cellvalue>,"^[0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3}$")
I have done till finding the value as 5.88.0 returns TRUE and 5.99 returns false, need to now append ".0" so 5.99 --> 5.99.0 and remove 'v' if found.
You can use a combination of functions, it may not be pretty, but it does the work
Replace any instance of v with an empty string using substitute, by making the content of the cell upper case, if we don't put UPPER(CELL) we could exclude any upper case V or lower case v(it will depend which function you use)
SUBSTITUTE(text_to_search, search_for, replace_with, [occurrence_number])
=SUBSTITUTE(UPPER(A1),"V","")
Look for cell missing the last block .xxx, you need to update a bit your regex to specified that the last group it's not present
^([0-9]{1}\.[0-9]{1,3} ( \.[0-9]{1,3}){0} )$
Using REGEXMATCH and IF we can then CONCATENATE the last group as .0
REGEXMATCH(text, regular_expression)
CONCATENATE(string1, [string2, ...])
=IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}(\.[0-9]{1,3}){0})$"),concatenate(A2,".0"), A2)
The last A2 will be replace with something similar than what we have until now, but before that we need to make small change in the regex, we want to look for the groups you specified were the last group it's present, that's your orignal regex, if it meets the regex it will put it in the cell, otherwise it will put INVALID, you can change that to anything you want it to be
^([0-9]{1}.[0-9]{1,3}.[0-9]{1,3})$
This it's the piece we are putting instead of the last A2
IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3})$"),substitute(upper(A2),"V",""),"INVALID")
With this the final code to put in your cell will be:
=IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}(\.[0-9]{1,3}){0})$"),concatenate(SUBSTITUTE(UPPER(A2),"V",""),".0"),IF(REGEXMATCH(substitute(upper(A2),"V",""),"^([0-9]{1}\.[0-9]{1,3}\.[0-9]{1,3})$"),substitute(upper(A2),"V",""),"INVALID"))
I have a string, and I want to extract, using regular expressions, groups of characters that are between the character : and the other character /.
typically, here is a string example I'm getting:
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
and so, I want to retrieved, 45.72643,4.91203 and also hereanotherdata
As they are both between characters : and /.
I tried with this syntax in a easier string where there is only 1 time the pattern,
[tt]=regexp(str,':(\w.*)/','match')
tt = ':45.72643,4.91203/'
but it works only if the pattern happens once. If I use it in string containing multiples times the pattern, I get all the string between the first : and the last /.
How can I mention that the pattern will occur multiple time, and how can I retrieve it?
Use lookaround and a lazy quantifier:
regexp(str, '(?<=:).+?(?=/)', 'match')
Example (Matlab R2016b):
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = regexp(str, '(?<=:).+?(?=/)', 'match')
result =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
In most languages this is hard to do with a single regexp. Ultimately you'll only ever get back the one string, and you want to get back multiple strings.
I've never used Matlab, so it may be possible in that language, but based on other languages, this is how I'd approach it...
I can't give you the exact code, but a search indicates that in Matlab there is a function called strsplit, example...
C = strsplit(data,':')
That should will break your original string up into an array of strings, using the ":" as the break point. You can then ignore the first array index (as it contains text before a ":"), loop the rest of the array and regexp to extract everything that comes before a "/".
So for instance...
'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh'
Breaks down into an array with parts...
1 - 'abcd'
2 - '45.72643,4.91203/Rou'
3 - 'hereanotherdata/defgh'
Then Ignore 1, and extract everything before the "/" in 2 and 3.
As John Mawer and Adriaan mentioned, strsplit is a good place to start with. You can use it for both ':' and '/', but then you will not be able to determine where each of them started. If you do it with strsplit twice, you can know where the ':' starts :
A='abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
B=cellfun(#(x) strsplit(x,'/'),strsplit(A,':'),'uniformoutput',0);
Now B has cells that start with ':', and has two cells in each cell that contain '/' also. You can extract it with checking where B has more than one cell, and take the first of each of them:
C=cellfun(#(x) x{1},B(cellfun('length',B)>1),'uniformoutput',0)
C =
1×2 cell array
'45.72643,4.91203' 'hereanotherdata'
Starting in 16b you can use extractBetween:
>> str = 'abcd:45.72643,4.91203/Rou:hereanotherdata/defgh';
>> result = extractBetween(str,':','/')
result =
2×1 cell array
{'45.72643,4.91203'}
{'hereanotherdata' }
If all your text elements have the same number of delimiters this can be vectorized too.
I have these strings:
fghs13412asdf
dfs234245gk
and want to return the position of the last numeric character, like so:
5
3
Perhaps there is something different in LibreOffice than Excel, where I'm seeing all the examples. Here's one that should be straightforward, and is returning an error.
Do you need the position of the first numeric character (as in the heading) or of the last one (as in the body of your question)?
If it's the first one, a simple SEARCH() function using regular expressions should to the trick, e.g. =SEARCH("([:digit:])";A1).
If it's the last one, counted from the start of the string, you can use a different regex (adapted from an answer in the OpenOffice forums by gerard24): =SEARCH("[0-9][^[0-9]]+$";A1).
If you need the position of the last numeric character, counted from the end of the string, just subtract the value calculated in step 2 from the LEN() of the entire string: =LEN(A1)-(SEARCH("[0-9][^[0-9]]+$";A1)).
You'll get a #VALUE! error if there's no numeric character, or if the last character of the input string is numeric. Note that whitespace in the string will be ignored:
Let's say I have a column which has values like:
foo/bar
chunky/bacon/flavor
/baz/quz/qux/bax
I.e. a variable number of strings separated by /.
In another column I want to get the last element from each of these strings, after they have been split on /. So, that column would have:
bar
flavor
bax
I can't figure this out. I can split on / and get an array, and I can see the function INDEX to get a specific numbered indexed element from the array, but can't find a way to say "the last element" in this function.
Edit:
this one is simplier:
=REGEXEXTRACT(A1,"[^/]+$")
You could use this formula:
=REGEXEXTRACT(A1,"(?:.*/)(.*)$")
And also possible to use it as ArrayFormula:
=ARRAYFORMULA(REGEXEXTRACT(A1:A3,"(?:.*/)(.*)$"))
Here's some more info:
the RegExExtract function
Some good examples of syntax
my personal list of Regex Tricks
This formula will do the same:
=INDEX(SPLIT(A1,"/"),LEN(A1)-len(SUBSTITUTE(A1,"/","")))
But it takes A1 three times, which is not prefferable.
You could do this too
=index(SPLIT(A1, "/"), COLUMNS(SPLIT(A1, "/"))-1)
Also possible, perhaps best on a copy, with Find:
.+/
(Replace with blank) and Search using regular expressions ticked.
You can try use this!
You've got the array of String, so you can acess the last element by length
String message = "chunky/bacon/flavor";
String[] outSplited = message.split("/");
System.out.println(outSplited[outSplited.length -1]);
I have a word, for example "ilikesamsung", and a dictionary of words, for example:
{"i","like","the","king","sam","sung","samsung"}
I want to calculate number of spaces in this word, if we break it into the dictionary words. In the above example, the string is broken into "i like samsung", and there are 2 spaces.
How can it be done?
Elaboration:
Problem has 2 parts-
Output Yes if this word can be broken.
Calculate number of spaces
What I have tried:
I have solved First part by Dynamic Programming.,
For the second part:
I have extended DP method to solve part2 . I took an array and I stored index at which word ends. arr[]={1,0,0,0,1,0,0,1,0,0,0,1}
this gives answer 3.
I want answer 2, for this array should be {1,0,0,0,1,0,0,0,0,0,0,1}.
Any advises?
You can use Dynamic Programming for this task:
f(0) = 0
f(i) = MIN { f(i-j) + (Dictionary.contais(s.substring(i-j,i)?1:INFINITY } for each j=1,...,i
The above finds the number of words in your string, so the final answer is f(#characters)-1
The idea is to do an "exhaustive search" - for a given new character - try to connect it to a word, and recursively invoke on what's left from the string.
This can be done pretty efficiently with DP techniques.