I have a card that I need it to show the Average of a column in a text style
this is what I have:
VAR Cnt = [CntPram]
var AllCnt = [CountAllPram]
RETURN
CntPram + " of " +[CountAllPram]
I end up getting this error
Cannot convert value 'of' of type Text to type Number.
How can I convert the numbers to make the card work?
You can use CONCATENATE for this.
Syntax
CONCATENATE(<text1>, <text2>)
This function joins two text strings into one text string. The joined items can be text, numbers or Boolean values represented as text, or a combination of those items. You can also use a column reference if the column contains appropriate values.
Read more on CONCATENATE
DAX does not recognize + as a concatenation operand for text srings. You can use nested CONCATENATE or join the strings using &:
Measure :=
[CntPram] & " of " & [CountAllPram]
Related
I'm new on python and I'm trying to convert a column of a dataframe with strings (like 10,000+ or 1,000+) with regex in order to eliminate characters (+ and ,) and then convert them into integer.
How can I do that?
I've tried with regex functions but it doesn't work
convert_installs = re.compile('(?P<amount>\d*).(?P<unit>\d*)')
is it correct for finding what I want to save?
enter image description here
df['columnname'] = df['columnname'].str.replace('([+,])', '', regex = True).astype('int')
Takes your column called columname, and replaces the + or , to nothing, then changes the type to integer.
So in my PostgreSQL 10 I have a column of type integer. This column represents a code of products and it should be searched against another code or part of the code. The values of the column are made of three parts, a five-digit part and two two-digit parts. Users can search for only the first part, the first-second or first-second-third.
So, in my column I have , say 123451233 the user searches for 12345 (the first part). I want to be able to return the 123451233. Same goes if the users also searches for 1234512 or 123451233.
Unfortunately I cannot change the type of column or break the one column into three (one for every part). How can I do this? I cannot use LIKE. Maybe something like a regex for integers?
Thanks
Consider to use simple arithmetic.
log(value)::int + 1 returns the number of digits in integer part of the value and using this:
value/(10^(log(value)::int-log(search_input)::int))::int
returns value truncated to the same digits number as search_input so, finally
search_input = value/(10^(log(value)::int-log(search_input)::int))::int
will make the trick.
It is more complex literally but also could be more efficient then strings manipulations.
PS: But having index like create index idx on your_table(cast(your_column as text)); search like
select * from your_table
where cast(your_column as text) like search_input || '%';
is the best case IMO.
You do not need regex functions. Cast the integer to text and use the function left(), example:
create table my_table(code int); -- or bigint
insert into my_table values (123451233);
with input_data(input_code) as (
values('1234512')
)
select t.*
from my_table t
cross join input_data
where left(code::text, length(input_code)) = input_code;
code
-----------
123451233
(1 row)
I would like to copy a certain string (out of a longer range of strings in one cell) and show it in a different cell with Google Sheets. This is what is in the initial cell A1:A :
"String 1","String 2","String 3"
In B1:B I'd like ONLY String 3, so without the "" and the other strings.
Is this possible with spreadsheets?
Or is there any other way of doing so?
Update
So the task is to get word inside double quotes. And the mathcing string is placed in the end of text.
You may use regular expressions to deal with that, the basic formula is:
=REGEXEXTRACT(A1,"([^""]+)""$")
This will give a word inside "" from text in cell A1 at the end of text.
For example:
some text...,"Thisthat","https://www.url.com/de/Thisthat"
gives https://www.url.com/de/Thisthat
You may also use arrayformula:
=ArrayFormula(REGEXEXTRACT(A1:A3,"([^""]+)""$"))
Please, read more about this functions here and here.
Old answer
if you want strings to be on their rows, use this formula in B1:
=ArrayFormula(if(A1:A = "String 3";A1:A;""))
If you have cells in A1:A, which contain 'string 3', and you want to match them too, use this:
=ArrayFormula(if(REGEXMATCH(A1:A , "String 3"),"String 3",""))
I'm trying to write a formula for Google Sheets which will convert Unicode characters with diacritics to their plain ASCII equivalents.
I see that Google uses RE2 in its "REGEXREPLACE" function. And I see that RE2 offers Unicode character classes.
I tried to write a formula (similar to this one):
REGEXREPLACE("público","(\pL)\pM*","$1")
But Sheets produces the following error:
Function REGEXREPLACE parameter 2 value "\pL" is not a valid regular expression.
I suppose I could write a formula consisting of a long set of nested SUBSTITUTE functions (Like this one), but that seems pretty awful.
Can any offer a suggestion for a better way to normalize Unicode letters with diacritical/accent marks in a Google Sheets formula?
[[:^alpha:]] (negated ASCII character class) works fine for REGEXEXTRACT formula.
But =REGEXREPLACE("público","([[:alpha:]])[[:^alpha:]]","$1") gives "pblic" as a result. So, I guess, formula doesn't know what exact ASCII character must replace "ú".
Workaround
Let's take the word públicē; we need to replace two symbols in it. Put this word in cell A1, and this formula in cell B1:
=JOIN("",ArrayFormula(IFERROR(VLOOKUP(SPLIT(REGEXREPLACE(A1,"(.)","$1-"),"-"),D:E,2,0),SPLIT(REGEXREPLACE(A1,"(.)","$1-"),"-"))))
And then make directory of replacements in range D:E:
D E
1 ú u
2 ē e
3 ... ...
This formula is still ugly, but more useful because you can control your directory by adding more characters to the table.
Or use Java Script
Also found a good solution, which works in google sheets.
This did it for me in Google Sheets, Google Apps Scripts, GAS
function normalizetext(text) {
var weird = 'öüóőúéáàűíÖÜÓŐÚÉÁÀŰÍçÇ!#£$%^&*()_+?/*."';
var normalized = 'ouooueaauiOUOOUEAAUIcC ';
var idoff = -1,new_text = '';
var lentext = text.toString().length -1
for (i = 0; i <= lentext; i++) {
idoff = weird.search(text.charAt(i));
if (idoff == -1) {
new_text = new_text + text.charAt(i);
} else {
new_text = new_text + normalized.charAt(idoff);
}
}
return new_text;
}
This answer doesn't require a Google App Script, and it's still fast, and relatively simple. It builds on Max's answer by providing a full lookup table, and it also allows for case-sensitive transliteration (normally VLOOKUP is NOT case-sensitive).
Here is a link to the Google Spreadsheet if you want to jump right into it. If you want to use your own sheet, you'll need to copy the TRANS_TABLE sheet into your Spreadsheet.
In the code snippet below, the source cell is A2, so you'd place this formula in any column on row 2. Using REGEXREPLACE AND SPLIT, we split apart the string in A2 into an array of characters, then USING ARRAYFORMULA, we do the following to EACH character in the array: First, the character is converted to its 'decimal' CODE equivalent, then matched against a table on the TRANS_TABLE sheet by that number, then using VLOOKUP, a character X number of columns over (the index value provided) on the TRANS_TABLE sheet (in this case, the 3rd column over) is returned. When all characters in the array have been transliterated, we finally JOIN the array of characters back into a single string. I provided examples with named ranges as well.
=iferror(
join(
"",
ARRAYFORMULA(
vlookup(
code(split(REGEXREPLACE($A2,"(.)", "$1;"),";",TRUE)),
TRANS_TABLE!$A$5:$F,3
)
)
)
,)
You'll note on the TRANS_TABLE sheet I made, I created 4 different transliteration columns, which makes it easy to have a column for each of your transliteration needs. To reference the column, just use a different index number in the VLOOKUP. Each column is simply a replacement character column. In some cases, you don't want any conversion made (A -> A or 3 -> 3), so you just copy the same character from the source Glyph column. Where you DO want to convert characters, you type in whatever character you want replaced (ñ -> n etc). If you want a character removed altogether, you leave the cell blank (? -> ''). You can see examples of the transliteration output on the data sheet in which I created 4 different transliteration columns (A-D) referencing each of the Transliteration tables from the TRANS_TABLE sheet for different use case scenarios.
I hope this finally answers your question in a fashion that isn't so "ugly." Cheers.
How can I use regular expression in excel ?
In above image I have column A and B. I have some values in column A. Here I need to move data after = in column B. For e.g. here in 1st row I have SELECT=Hello World. Here I want to remove = sign and move Hello world in column B. How can I do such thing?
Stackoverflow has many posts about adding regular expressions to Excel using VBA. For your particular example, you would need VBA to actually move a substring from one cell to another.
If you simply want to copy the substring, you can do so easily using the MID function:
=IFERROR(MID(A1,FIND("=",A1)+1,999),A1)
I used 999 to ensure that enough characters were grabbed.
IFERROR returns the cell as-is if an equals sign is not found.
To return the portion of string before the equals sign, do this:
=LEFT(A1,FIND("=",A1&"=")-1)
In this case, I appended the equals sign to A1, so FIND won't return an error if not found.
You can use the Text to Column functionality of MS-Excel giving '=' as delimiter.
Refer to this link:
Chop text in column to 60 charactersblocks
You can simply use Text to Column feature of excel for this:
Follow the below steps :
1) Select Column A.
2) Goto Data Tab in Menu Bar.
3) Click Text to Column icon.
4) Choose Delimited option and do Next and then check the Other options in delimiter and enter '=' in the entry box.
5) Just click finish.
Here are URL for Text to Column : http://www.excel-easy.com/examples/text-to-columns.html