Regex Extract All using pig - regex

This is my sample data (attributes) to match with:
[87027^适用年龄#619081^12-36个月,108389^国产/进口#619192^国产,108376^包装单位#619113^盒装,12210^分类#121491^牛奶粉,108385^奶源产地#619164^澳洲/新西兰,115919^配方#651827^常规配方奶粉,12212^段位#121499^3段,108380^净含量#619152^2kg以上]
[108385^奶源产地#619183^其它 ,12212^段位#121499^3段,87027^适用年龄#619081^12-36个月,108376^包装单位#619120^桶装,108389^国产/进口#619196^进口,12210^分类#121491^牛奶粉,115919^配方#651827^常规配方奶粉,12241^价格#121803^200-299元,108380^净含量#619147^800-1000g]
[87027^适用年龄#619081^12-36个月,108389^国产/进口#619192^国产,108376^包装单位#619113^盒装,12210^分类#121491^牛奶粉,108385^奶源产地#619164^澳洲/新西兰,115919^配方#651827^常规配方奶粉,12212^段位#121499^3段,108380^净含量#619152^2kg以上]
So I'm trying to match with all the values in between the numbers. So for the first data line, I would want to match with the value between 87027 and 619081, and the value between 619081 and 108389 and so on for every value.
Here is my code so far:
data = LOAD 'SHORT.txt' USING PigStorage() AS (number:chararray, user:chararray, item:chararray, attributes:chararray);
B = FOREACH data GENERATE REGEX_EXTRACT_ALL (attributes, '\\^(.*?)[#|,|\\]]');
DUMP B;
Please let me know if you there is any other information needed. Thanks!

This pattern captures what you describe:
\d+?(\^.+?[,#\]])
Example here: https://regex101.com/r/pOPWwY/2
Looks like Hadoop needs double-escaping of backslashes, so bear that in mind when testing the above.
If want to exclude the markers , # ], then:
\d+?(\^.+?)[,#\]]

Related

regular expression search backwards, How to deal with words with and without?

I tested by https://regexr.com/
There two sample words.
BOND_aa_SB1_66-1.pdf
BOND_bb_SB2.pdf
I want to extract SB1, SB2 from each sample.
but my regular expression is not perfect.
It is working
(?<=BOND_.*_).*
But It is difficult to write the following.
I try
(?<=BOND_.*_).*(?=(_|\.))
But first sample result is 'SB1_66-1'
I just want to extract SB1
sb1 The following may or may not exist. if there is content, it can be separated by starting with _.
How should I fix it?
To extract the third underscore-separated term, we can use re.search as follows:
inp = ["BOND_aa_SB1_66-1.pdf", "BOND_bb_SB2.pdf"]
output = [re.search(r'^BOND_[^_]+_([^_.]+)', x).group(1) for x in inp]
print(output) # ['SB1', 'SB2']
s = "BOND_aa_SB1_66-1.pdf BOND_bb_SB2.pdf"
(re.findall(r'(SB\d+)', s))
['SB1', 'SB2']

Google apps scripts: regexmatch based on two criteria including a passed variable

I’m using SpreadsheetApp.newFilterCriteria().whenFormulaSatisfied() to filter a sheet, but I’m not getting the regexmatch() function setup properly. I need to filter column A based on two criteria:
must contain “ak1” or “ak2”
And, must contain myVar which is a variable that will be passed to the script.
I know how to write the regexmatch statement to get criteria 1 (regexmatch(A:A, "(?i)ak1|ak2")), but I’m getting stuck on how to combine that with criteria 2.
My best attempt so far:
var regex = new RegExp( myVar, 'g' );
var crit = '=regexmatch(A:A, "^(?=.*ak1|ak2|)(?=.*myVar)")'
Edited to add example:
If col A contains
Chr_rq_ak1
Chr_rq_ak2
Chr_rq_ak6
Chr_gv_ak1
I would expect the following to be visible and the remainder hidden:
Chr_rq_ak1
Chr_rq_ak2
You would use AND:
=AND(REGEXMATCH(A2,"(?i)ak(?:1|2)"),REGEXMATCH(A2,"myVar"))
If you want to use the alternation you have to group them(?=.*(?:k1|ak2)) and leave out the | at the end.
If you have only a single char following, you could also use a character class.
^(?=.*ak[12])(?=.*myVar)
Regex demo

Extract string of numbers from URL using regex PIG

I'm using PIG to generate a list of URLs that have been recently visited. In each of the URLs, there is a string of numbers that represents the product page visited. I'm trying to use a regex_extract_all() function to extract just the string of numbers, which vary in length from 6-8. The string of digits can be found directly after jobs2/view/ and usually ends with +&cd but sometimes they may end with ).
Here are a few example URLs:
(http://a.com/search?q=cache:QD7vZRHkPQoJ:ca.xyz.com/jobs2/view/17069404+&cd=1&hl=en&ct=clnk&gl=ca)
(http://a.com/search?q=cache:G9323j2oNbAJ:ca.xyz.com/jobs2/view/5977065+&cd=1&hl=en&ct=clnk&gl=ca)
(http://a.com/search?q=cache:aNspmG11qAJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=clnk&gl=hk)
(http://a.com/search?q=cache:aNspmG11AJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=clnk&gl=hk)
(http://a.com/search?q=cache:aNspmG11qAJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=cl k&gl=hk)
Here is the current regex I am using:
J = FOREACH jpage GENERATE FLATTEN(REGEX_EXTRACT_ALL(TEXTCOLUMN, '\/view\/(\d+)\+\&')) as (output:chararray)
I have also tried other forms such as:
'[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]', 'view.([0-9]+)', 'view\/([\d]+)\+',
'[0-9][0-9][0-9]+', and
'[0-9][0-9][0-9]*'; none of which work.
Can anybody assist here or have another way of going about it?
Much appreciated,
MM
Reason for"Unexpected character 'D'" is, you need to put double backslash instead of single backslash. eg just replace [\d+] to [\\d+]
Here your solution, please validate all your inputs strings
input.txt
http://a.com/search?q=cache:QD7vZRHkPQoJ:ca.xyz.com/jobs2/view/17069404+&cd=1&hl=en&ct=clnk&gl=ca
http://a.com/search?q=cache:G9323j2oNbAJ:ca.xyz.com/jobs2/view/5977065+&cd=1&hl=en&ct=clnk&gl=ca
http://a.com/search?q=cache:aNspmG11qAJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=clnk&gl=hk
http://a.com/search?q=cache:aNspmG11AJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=clnk&gl=hk
http://a.com/search?q=cache:aNspmG11qAJ:hk.xyz.com/jobs2/view/16988928+&cd=2&hl=zh-TW&ct=clk&gl=hk
http://a.com/search?q=cache:aNspmG11qAJ:hk.xyz.com/jobs2/view/16988928)=2&hl=zh-TW&ct=clk&gl=hk
http://webcache.googleusercontent.com/search?q=cache:http://my.linkedin.com/jobs2/view/9919248
Updated Pigscript:
A = LOAD 'input.txt' as line;
B = FOREACH A GENERATE REGEX_EXTRACT(line,'.*/view/(\\d+)([+|&|cd|)?]+)?',1);
dump B;
(17069404)
(5977065)
(16988928)
(16988928)
(16988928)
(16988928)
I'm not familiar with PIG, but this regex will match your target:
(?<=/jobs2/view/)\d+
By using a (non-consuming) look behind, the entire match (not just a group of the match) is your number.

How to include 2 words within Regex and result must be based on only those 2 words VB.NET

I would like to know how to include only 2 or more keywords within a Regex. and ending results should only show those words defined, not only one word.
What I currently have works with multiple keywords but I want it to use BOTH words not either one of the other.
For example:
Dim pattern As String = "(?i)[\t ](?<w>((arma)|(crapo))[a-z0-9]*)[\t ]"
Now the code works fine by including 'arma' or 'crapo'. I only want it to include BOTH 'arma' AND 'crapo' otherwise do not show any results.
Dealing with finding certain keywords within a PDF document and I only want to be shown results if the PDF document includes BOTH 'arma' and 'crapo' (Works fine by showing results for 'arma' OR 'crapo' I want to see results based on 'arma' AND 'crapo'.
Sorry for sounding so repetitive.
Edit: Here is my code. Please read comment.
Dim filesz() As String = GetPatternedFiles("c:\temp\", New String() {"tes*.pdf", "fes*.pdf", "Bas*.pdf"})
'The getpatterenedfiles is a function" also gettextfromPDF is another function.
For Each s As String In filesz
Dim thetext As String = Nothing
Dim pattern As String = "(?i)[\t ](?<w>(crapo)|(arma)[a-z0-9]*)[\t ]"
thetext = GetTextFromPDF(s)
For Each m As Match In Regex.Matches(thetext, pattern)
ListBox1.Items.Add(s)
Next
Next
You can use this regex:
\barma\b.*?\bcrapo\b|\bcrapo\b.*?\barma\b
Working demo
The idea is to match arma whatever crapo or crapo whatever arma and use word boundaries to avoid words like karma.
However, if you want to match karma or crapotos as you asked in your comment you can use:
arma.*?crapo|crapo.*?arma

R: replacing special character in multiple columns of a data frame

I try to replace the german special character "ö" in a dataframe by "oe". The charcter occurs in multiple columns so I would like to be able to do this all in one by not having to specify individual columns.
Here is a small example of the data frame
data <- data.frame(a=c("aö","ab","ac"),b=c("bö","bb","ab"),c=c("öc","öb","acö"))
I tried :
data[data=="ö"]<-"oe"
but this did not work since I would need to work with regular expressions here. However when I try :
data[grepl("ö",data)]<-"oe"
I do not get what I want.
The dataframe at the end should look like:
> data
a b c
1 aoe boe oec
2 ab bb oeb
3 ac ab acoe
>
The file is a csv import that I import by read.csv. However, there seems to be no option to change to fix this with the import statement.
How do I get the desired outcome?
Here's one way to do it:
data <- apply(data,2,function(x) gsub("ö",'oe',x))
Explanation:
Your grepl doesn't work because grepl just returns a boolean matrix (TRUE/FALSE) corresponding to the elements in your data frame for which the regex matches. What the assignment then does is replace not just the character you want replaced but the entire string. To replace part of a string, you need sub (if you want to replace just once in each string) or gsub (if you want all occurrences replaces). To apply that to every column you loop over the columns using apply.
If you want to return a data frame, you can use:
data.frame(lapply(data, gsub, pattern = "ö", replacement = "oe"))