Both FindNoCase and Find funcions are returning 0 values for all cases. I'm working on Coldfusion 9.
<cfoutput>#Find("aaInternationalbb", "International")#</cfoutput> ->o/p:0
<cfoutput>#Find("aalbb", "International")#</cfoutput> ->o/p:0
<cfoutput>#FindNoCase("aaInternationalbb", "International")#</cfoutput> ->o/p:0
<cfoutput>#FindNoCase("aalbb", "International")#</cfoutput> ->o/p:0
Please let me know how to make this work. Thanks in advance
This is not working because you have given the parameters in wrong order. The syntax for each of find() and findNoCase() are
FindNoCase(substring, string [, start ])
Find(substring, string [, start ])
So you have to try like:
<cfoutput>#Find( "International","aaInternationalbb")#</cfoutput>
<cfoutput>#Find( "International","aalbb")#</cfoutput>
<cfoutput>#FindNoCase("International","aaInternationalbb" )#</cfoutput>
<cfoutput>#FindNoCase("International","aalbb")#</cfoutput>
This will give output as: 3 0 3 0
findNoCase and Find will expect ( stringtosearch, stringtosearchfrom [,start])
Related
I want to make the following transformation to a set of datas in my google spreadsheets :
6 views -> 6
73K views -> 73000
3650 -> 3650
163K views -> 163000
1.2K views -> 1200
52.5K -> 52500
All the datas are in a column and depending on the case I need to apply a specific transformation.
I tried to put all the regex in one formula but I failed. I always had a case over two regular expressions etc.
Anyaway I end up making these regex one case by one case in different columns. It works fine but I feel like it could slowdown the sheet since I except a lot of data coming into this sheet.
Here is the sheet : spreadsheet
Thank you for your help !
Use regexreplace(), like this:
=arrayformula(
iferror( 1 /
value(
regexreplace(
regexreplace(trim(A2:A), "\s*K", "e3"),
" views", ""
)
)
^ -1 )
)
See your sample spreadsheet.
replace 'views' using regex: /(?<=(\d*\.?\d+\K?)) views/gi
To replace 'K' with or without decimal value, first, detect K then replace K with an empty string and multiply by 1000.
use call back function as:
txt.replace(/(?<=(\d*\.?\d+\K?)) views/gi, '').replace(/(?<=\d)\.?\d+K/g, x => x.replace(/K/gi, '')*1000)
code:
arr = [`6 views`,
`73K views`,
`3650`,
`163K views`,
`1.2K views`,
`52.5K`];
arr.forEach(txt => {
console.log(txt.replace(/(?<=(\d*\.?\d+\K?)) views/gi, '').replace(/(?<=\d)\.?\d+K/g, x => x.replace(/K/gi, '')*1000))
})
Output:
6
73000
3650
163000
1200
52500
Say your inputs are in column A. Empty cells allowed. In any other column,
=arrayformula(if(A2:A<>"",value(substitute(substitute(A2:A," views",""),"K","e3")),))
works.
Adjust the range A2:A as needed.
Also note that non-empty cells with empty strings are ignored.
Basically, since Google Sheet's regex engine doesn't support look around, it is more efficient to take advantage of the rather strict patterns in your application and use substitute() instead.
I have a spreadsheet where I look if the data (website) already exist on the master sheet.
=if(countif(importrange("Spreadsheet Key","Leads!N:N"),K2)>0,"COMPANY EXISTS!","")
But the above formula is not dynamic enough. If there are companies with co.uk and on the master sheet if it's registered under .com, it won't show "COMPANY EXISTS!"
So I changed to the formula to look for works after before and after "." on a website.
=ARRAYFORMULA(REGEXEXTRACT(UNIQUE(SUBSTITUTE(importrange("Spreadsheet Key","Leads!N:N"),"www.","")), "([0-9A-Za-z-]+)\."))
But it's not working if I try to incorporate with if and countif.
=if(COUNTIF(ARRAYFORMULA(REGEXEXTRACT(SUBSTITUTE(importrange("Spreadsheet Key","Leads!N:N"),"www.",""), "([0-9A-Za-z-]+)\."),L2:L)>0,"Company Exist!",""))
It shows 'Wrong number of arguments to IF. Expected between 2 and 3 arguments, but got 1 arguments'
Can anyone help me out on where I am making the mistake?
Spreadsheet link- https://docs.google.com/spreadsheets/d/1La3oOWiM5KpzRY0MLLEUQC25LzDuQlqTjgFp-VlS8Bo/edit#gid=0
Edited: Made a mistake beforehand, didn't specify on that cell its looking against
Your formula had a typo. You are not closing the Arrayformula and the Countif correctly (the array formula closing parenthesis should go before the , of the count if). So change this:
=if(COUNTIF(ARRAYFORMULA(REGEXEXTRACT(SUBSTITUTE(importrange("Spreadsheet Key","Leads!N:N"),"www.",""), "([0-9A-Za-z-]+)\."),L2:L)>0,"Company Exist!",""))
To this:
=if(COUNTIF(ARRAYFORMULA(REGEXEXTRACT(SUBSTITUTE(IMPORTRANGE("Spreadsheet Key","Leads!N:N"),"www.",""), "([0-9A-Za-z-]+)\.")),L3:L)>0,"Company Exist!","")
I hope this has helped you. Let me know if you need anything else or if you did not understood something. :)
try:
=ARRAYFORMULA(IFNA(IF(IFNA(REGEXEXTRACT(SUBSTITUTE(IMPORTRANGE(
"1bnz7Y_xVN9Jo80aCBBeMBMJBnMDHkbZQUWnmL20CRi8", "Leads!N:N"),
"www.", ), "([0-9A-Za-z-]+)\."))>0, "Company Exist!", )))
def max(*args):
return sorted(args[-1])
print max([1,3],[1,4],[2,1])
>>>[1,2]
I am getting (1,2) as answer which is not even in the list i supplied.I understand it has something to do with [-1].following are the values returned when i tried with 0 ,1 ,2 in place of -1
0-->1,3
1-->1,4
2-->1,2
Can someone explain what is happening
sorry guys. my bad . I found out that correct code should be:
def max(*args):
return sorted(args)[-1]
I have a column in data frame which ex df:
A
0 Good to 1. Good communication EI : tathagata.kar#ae.com
1 SAP ECC Project System EI: ram.vaddadi#ae.com
2 EI : ravikumar.swarna Role:SSE Minimum Skill
I have a list of of strings
ls=['tathagata.kar#ae.com','a.kar#ae.com']
Now if i want to filter out
for i in range(len(ls)):
df1=df[df['A'].str.contains(ls[i])
if len(df1.columns!=0):
print ls[i]
I get the output
tathagata.kar#ae.com
a.kar#ae.com
But I need only tathagata.kar#ae.com
How Can It be achieved?
As you can see I've tried str.contains But I need something for extact match
You could simply use ==
string_a == string_b
It should return True if the two strings are equal. But this does not solve your issue.
Edit 2: You should use len(df1.index) instead of len(df1.columns). Indeed, len(df1.columns) will give you the number of columns, and not the number of rows.
Edit 3: After reading your second post, I've understood your problem. The solution you propose could lead to some errors.
For instance, if you have:
ls=['tathagata.kar#ae.com','a.kar#ae.com', 'tathagata.kar#ae.co']
the first and the third element will match str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i])
And this is an unwanted behaviour.
You could add a check on the end of the string: str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i]+r'(?:\s|$)')
Like this:
for i in range(len(ls)):
df1 = df[df['A'].str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i]+r'(?:\s|$)')]
if len(df1.index != 0):
print (ls[i])
(Remove parenthesis in the "print" if you use python 2.7)
Thanks for the help. But seems like I found a solution that is working as of now.
Must use str.contains(r'(?:\s|^|Ei:|EI:|EI-)'+ls[i])
This seems to solve the problem.
Although thanks to #IsaacDj for his help.
Why not just use:
df1 = df[df['A'].[str.match][1](ls[i])
It's the equivalent of regex match.
I want to display an output with Test = 0 using Fortran, I tried to use:
'WRITE(11,*) 'Test =' testdata'
Assuming 11 is correct and testdata is a parameter that is being calculated.
I wasn't able to get the output and there was an error.
Anyone have any idea why it is so?
Try inserting a comma and deleting the apostrophes:
WRITE(11,*) 'Test =', testdata
If you had reported what the error message you saw was I might have made this answer more apposite.