I have a Regex table to identify classifications for different pages.
It works fine but the output value is somewhat scribbeled.
There might be character or two prepended or appended in the final Regex Table output value.
If the page is identified as My Class One (assigned output value) then the final value is something like dMy Class Onexz.
Regex table consists of piped values like asdsd|zxcz|zxzc and output field has readable value such as My Class One.
The final data is now a bit of a mess due these extra characters. Why is this happening? What to do?
EDIT: I don't know that what value a image brings since it has the same information as described here.
Related
I need to align text in an ALERT STRING column with the row identified by number in an ID ROW column.
Additionally, I need to also align the same ALERT STRING text with the same ID ROW number AND with the ID matching that embedded in a string in the TEXT WITH ID column. (This double-check will sometimes be necessary with the real-world data.)
So far, I've only figured out how to align the ALERT STRING with the ID matching that embedded in the TEXT WITH ID column:
=LOOKUP(2,1/SEARCH(A2,$F$2:$F$11),$G$2:$G$11)
I appreciate any help folks can offer. You can find an editable copy of the workbook here:
https://1drv.ms/x/s!ArQ7Kw6ayNMY2zktTW3pDCbMmJZ_
UPDATE: Nayan provided a solution to the first part of this question (please see answer below). I'm still trying to work out a formula for the column D part of this question, in which the row reference shown in column E is combined with a match of the ID shown in column A with its corresponding value in one of the text strings in column F.
The best I've been able to come up with so far is a formula with a high failure rate:
=INDEX($G$2:$G$11,MATCH(ROW(D2),$E$2:$E$11,MATCH("*"&A2&"*",$F$2:$F$11,0)))
Any help with this part of the question will be greatly appreciated.
ROW([reference])
Returns the row number of a reference
E.g.: Row(B2) returns 2. If nothing provided like ROW() will also
return row number based on position of cell where it is called.
VLOOKUP(loolup_value, table_array, col_index_num, [range_lookup])
Looks for a value in the leftmost column of a table, and then returns a value in the same row from a column you specify (col_index_num)
By default - the table must be sorted in an ascending order.
Try this:
=VLOOKUP(ROW(B2),$E$2:$G$11,3,FALSE)
INDEX(array, row_num, [column_num]) INDEX(reference, row_num,
[column_num], [area_num])
Returns a value or reference of the cell at the intersection of a particular row and column, in a given range.
In this case, you have to get row_num with MATCH function.
MATCH(lookup_value, lookup_array, [match_type])
Returns a relative position of an item in an array that matches a specified value in a specified order.
match_type: 1 (Less than), 0 (Exact match), -1 (Greater than)
Try this:
=INDEX($G$2:$G$11,MATCH(ROW(B2),$E$2:$E$11,0))
Identity Data with Multiple Criteria Condition using MATCH()
=INDEX($G$2:$G$11,MATCH(1, (ROW(D2) = $E$2:$E$11) * (ISNUMBER(SEARCH(A2, $F$2:$F$11))),0))
References:
https://exceljet.net/excel-functions/excel-vlookup-function
https://exceljet.net/excel-functions/excel-index-function
https://exceljet.net/formula/index-and-match-with-multiple-criteria
This is the formula I was looking for in column D:
=INDEX($G$2:$G$11,MATCH(ROW(D2)&"*"&A2&"*",INDEX($E$2:$E$11&$F$2:$F$11,),0))
You can see it working here.
Nayan provided a great deal of help with answering this question, so I will mark his answer as the accepted solution.
Syeda Fahima Nazreen provided the example I referenced to figure out the formula shown above.
Reference:
Nested Excel Formula with Two INDEX Functions and a MATCH Function with Multiple Criteria
forest area to the I want to add a column name (say ForestAreaPerPopn) to find the ratio of forest area to the population(represented by variable Total below) residing. The data contains the following variables and their values.
How can I add a column named ForestAreaPerPopn in Table****ForestAreaPerPop (shown below) so that the column contains the data calculated as ratio of forest area to Total.
Too long for a comment.
You have a couple of problems. First, your column names have spaces and other special characters. This is allowed but creates all kinds of problems later. I suggest you do something like:
colnames(ForestAreaPerPop) <- gsub(' |\\(|\\)', '_', colnames(ForestAreaPerPop))
This will replaces any spaces, left or right parens in the colnames with '_'.
Then, something like:
ForestAreaPerPop$n <- with(ForestAreaPerPop, Forest_Area_in_ha/Total)
should give you what you want.
Some advice: long table names and column names may seem like a good idea, but you will live to regret it. Make them short but meaningful (easier said than done).
I have tens of thousands of rows of unstructured data in csv format. I need to extract certain product attributes from a long string of text. Given a set of acceptable attributes, if there is a match, I need it to fill in the cell with the match.
Example data:
"[ROOT];Earrings;Brands;Brands>JeweleryExchange;Earrings>Gender;Earrings>Gemstone;Earrings>Metal;Earrings>Occasion;Earrings>Style;Earrings>Gender>Women's;Earrings>Gemstone>Zircon;Earrings>Metal>White Gold;Earrings>Occasion>Just to say: I Love You;Earrings>Style>Drop/Dangle;Earrings>Style>Fashion;Not Visible;Gifts;Gifts>Price>$500 - $1000;Gifts>Shop>Earrings;Gifts>Occasion;Gifts>Occasion>Christmas;Gifts>Occasion>Just to say: I Love You;Gifts>For>Her"
Look up table of values:
Zircon, Diamond, Pearl, Ruby
Output:
Zircon
I tried using the VLOOKUP() function, but it needs to match an entire cell and works better for translating acronyms. Haven't really found a built in function that accomplishes what I need. The data is totally unstructured, and changes from row to row with no consistency even within variations of the same product. Does anyone have an idea how to do this?? Or how to write an OpenOffice Calc function to accomplish this? Also open to other better methods of doing this if anyone has any experience or ideas in how to approach this...
ok so I figured out how to do this on my own... I created many different columns, each with a keyword I was looking to extract as a header.
Spreadsheet solution for structured data extraction
Then I used this formula to extract the keywords into the correct row beneath the column header. =IF(ISERROR(SEARCH(CF$1,$D769)),"",CF$1) The Search function returns a number value for the position of a search string otherwise it produces an error. I use the iserror function to determine if there is an error condition, and the if statement in such a way that if there is an error, it leaves the cell blank, else it takes the value of the header. Had over 100 columns of specific information to extract, into one final column where I join all the previous cells in the row together for the final list. Worked like a charm. Recommend this approach to anyone who has to do a similar task.
Here is a challenge for you: I was trying to make a tic tac toe based on R. First, the players have to configure putting in the name of the players, and the game should check if the name exists in a file called "Players.txt" (if not, the game will create one), if the name exists, the game will ask for a new one. The last part of the game is that the game should record all the punctuation of the players (each gambling chip used will subtract 5 points of 100 that the player has at the beginning of the game). The problem is when a player wins, the game shows the following error: "Error in table[location_name1, 3]: Incorrect number of dimension in R".
A vector can either be atomic or a list. Atomic vectors can only contain elements of one and the same data type. That means, you are "accidentally" creating a list with
vector=c(win,name1,name2,table)
with the result that each column of the data frame should become an entry.
You can solve it with
vector <- list(win, name1, name2, table)
vector is still a list but now it has the format I believe you want.
Having done that you still get errors. The reason is that these assignments fail.
location_name1=which(grepl(name1,table$gamers))
location_name2=which(grepl(name2,table$gamers))
They return an empty vector because earlier in the code you set win=vector[1]... table=vector[4]. Since vector is now a list, you have to subset it accordingly. That means you have to chance the statements to table=vector[[4]].
Now you are going to get another problem. The reason is that you treat the columns table$scores as text. When you read the data you need to make sure that this columns is not interpreted as text. You also have to eliminate all statements that coerce the column into text. Otherwise table[location_name1,3]=table[location_name1,3]+pointsx will obviously fail because you cannot add a number to a string.
For example, you coerce the column into a character column with this statement:
name1 <- data.frame(gamers=name1,games="1",scores="100")
games and scores are strings not numbers. Another example is the assigment after reading the table from the file. You can make sure that scoresare numeric by doing this.
scores <- as.numeric(table[,3])
Please get familiar with Rstudio debugging capabilities (https://support.rstudio.com/hc/en-us/articles/205612627-Debugging-with-RStudio). This way you can go through your code line by line and check consequences of each assignment to the data frame.
I am trying to compare two lists of DateTime values in a Google Sheets. I want to highlight all DateTimes in the first list which are also present in the second list.
I already tried to use MATCH(), COUNTIF(), FILTER() together with COUNTA() or other approaches. However, although the values in both lists are basically copies of each other (just with some missing values in the second list), no matches will be returned. All are "true" DateTime values which can be formatted by using any of the date and time formatting options.
The MATCH example:
=ISNUMBER(MATCH(A2,INDIRECT("OTHERSHEET!$A$2:$A"),0))
I assume that there might be differences in the DateTime interpretation in the comparison why I also tried to use N() without success. I will always get an error like Did not find value '43297.75867' in MATCH evaluation. respectively FALSE after the ISNUMBER().
If I just do something like =N(A1)=N(OTHERSHEET!A1) with matching DateTimes, I get TRUE.
Same principle, but shorter:
=MATCH(A1,INDIRECT("OTHERSHEET!A:A"),0)
custom formula: =ISNUMBER(MATCH(A1,INDIRECT("OTHERSHEET!$A$1:$A"),0))