xmatch | Simple pattern matching for ES6 | Script Programming library
kandi X-RAY | xmatch Summary
kandi X-RAY | xmatch Summary
Simple pattern matching for ES6 (no transpilation!).
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- Creates an iterator over the next element in the array of patterns .
xmatch Key Features
xmatch Examples and Code Snippets
Community Discussions
Trending Discussions on xmatch
QUESTION
With a very simple function that returns a dynamic range in A7:
...ANSWER
Answered 2021-Jun-02 at 21:03If your List is only Text, you could use
QUESTION
ANSWER
Answered 2021-Jun-02 at 19:07Use COUNTA
and the spilled range operator #
:
In C10
:
QUESTION
Consider a very basic table:
...ANSWER
Answered 2021-May-20 at 18:13One can use LET
to avoid the repeat:
QUESTION
I'm using a modified match function since years which worked fine. But suddenly I get runtime error 13 for no reason. This function is called two times in the actual process before the error pops up. The first time everything works fine, the second time I get the error. Here's the code:
...ANSWER
Answered 2021-May-13 at 19:03TL;DR:
Change For Each xMTcell In Direction_Range
to For Each xMTcell In Direction_Range.Cells
.
Some debugging and general tips (summarizing the comments and your feedback in them):
Dim xMTcell
- make that aRange
.- Use
Debug.Print
(Control+G to bring up the Immediate Window and inspect the output). Debug.Print VarType(xMTcell.Value)
returns8204
: Per theVarType
docs, that means thatxMTcell.Value
is avbArray
ofvbVariant
s (8192 + 12 = 8204).Debug.Print VarType(Find_Value_or_String)
returns8
: Again per theVarType
docs, this meansFind_Value_or_String
is aString
.- The type mismatch is because you can't compare a
String
to an array. - The fact that
xMTCell.Value
is an array points toxMTCell
being a multi-cell range, not a single cell. - ... which is verified by the output of
Debug.Print xMTcell.Address
being a multi-cell range. - Most likely the issue is that you passed a
Row
orColumn
as yourDirection_Range
, i.e. you usedRows
orColumns
to return a range. When looping over aRow
orColumn
, you need to specify that you're looping over the individual cells.
QUESTION
I think I made a breakthrough with "ultra-dynamic" ranges. That's how I call them because not only they are dynamic; but also they refer to different ranges based on the cell address in which they are written in! Now I need to advance this breakthrough even further. Those of you using dynamic ranges in Excel, especially the dynamic-range-gurus, will be thrilled to read below and can possibly help in this advancement:
Disclaimer: If you are not familiar with dynamic ranges do not attempt to read below!
Background: Our sheet has cells with calculations on top and a pivot table below. Each cell above refers to the pivot table cells below in the same column. The first column of the pivot table (titled "Row Labels" by default) is sorted from top to bottom in descending order. Each of the next columns has the result of a different test. At one point somewhere in the middle rows of the pivot table there is a "marker line" that separates the top part of the pivot from the bottom one. Let's call the top part "Uppers" and the bottom part "Downers". Let's call both parts together "Population". Population is a non-contiguous range because the "marker line" that separates the Uppers from the Downers intervenes.
For each cell above the pivot there are calculations for the pivot column exactly below that need to refer to the Uppers or Downers or Population of the column itself.
Previously, my formulas in all of the cells above were a repetition of something like this below:
...ANSWER
Answered 2021-Apr-28 at 17:20The correct formula that defines an ultra-dynamic-range for replacing the 'OFFSET' formula:
=OFFSET($A$79,$B$5+1,COLUMN()-1,$B$6,1)
Is:
=INDEX($A:$XX,ROW($A$79)+$B$5+1,COLUMN()):INDEX($A:$XX,ROW($A$79)+$B$7,COLUMN())
- Where in
B7
is the number of the last row of the pivot. (=Something likeB5+B6
plus/minus 1 or 2 - test it for your case)
If you Define a Name of a Dynamic Range with this ultra-dynamic formula, it will adapt to give you different ranges depending on the location of the cell where you copied it to!! It will always give you the same parallel range in your column and will produce different results in different columns! I tested it and it works perfectly, plus it made my calculations lightning faster since INDEX
is non-volatile (as opposed to OFFSET
).
One more tip for the example above: I also tried nested range names and they work! For example I Defined the Range for ColPopulation as:
=(ColUppers,ColDowners)
Note that this is an ultra-dynamic-nested-range-name! Of course, it can work only for simple functions (such as =MAX(ColPopulation)
) and will not work in functions such as SUMPRODUCT
that need contiguous ranges. Still, it is a very useful thing to know that you can Define Ranges by adding other range names with commas!
Great help from all involved! Thanks a lot!
QUESTION
What array formula will return values which don't appear in another list?
Example:
Cells named ShortList
contain (one word per cell):
ANSWER
Answered 2021-Apr-19 at 07:52i think i got it. This returns the expected result in my question.
QUESTION
I am facing an issue to find all strings which have first few (>=2) characters repeated twice in a string in R language.
E.g
The strings should select out
(1) allochirally ------> first 3 characters 'all' repeated twice in string
(2) froufrou ------> first 4 characters 'frou' repeated twice in string
(3) undergrounder ------> first 5 characters 'under' repeated twice in string
The strings should NOT select out
(1) gummage ------> even first character 'g' repeated twice, but only 1 character, not match condition as >=2 first characters
(2) hypergoddess ------> no first few characters repeated twice
(3) kgashga ------> even 'ga' repeated twice, but not including the first character 'k', not match condition which require including the first character
Heard about backreference
(e.g \b or \w) might be helpful, but still not able to figure out, could you help to figure out ?
Note: I see there is a function as xmatch <- str_extract_all(x, regex) == x
as the method to use, the str_extract_all
from library(stringr)
ANSWER
Answered 2020-Feb-13 at 11:11Use grepl
:
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