I want to create list of all occurences of "x" string in range. This is my sheet:
And I want to search all occurences and list them and give proper names:
For example for G2, I want "Beret Grey" string as result. I think that I need to use array formula or something like that.
Let me first preface this that vba would be much more robust, but this formula will get you there. It may be slow as it is an array type formula and is doing a lot of calculations. These calculations only expound exponentially as the number of cells with them in it increases:
=IFERROR(INDEX(A:A,AGGREGATE(15,6,ROW($B$2:$G$7)/($B$2:$G$7="x"),ROW(1:1))) & " " & INDEX($1:$1,AGGREGATE(15,6,COLUMN(INDEX(A:G,AGGREGATE(15,6,ROW($B$2:$G$7)/($B$2:$G$7="x"),ROW(1:1)),0))/(INDEX(A:G,AGGREGATE(15,6,ROW($B$2:$G$7)/($B$2:$G$7="x"),ROW(1:1)),0)="x"),ROW(1:1)-COUNTIF($B$1:INDEX(G:G,AGGREGATE(15,6,ROW($B$2:$G$7)/($B$2:$G$7="x"),ROW(1:1)) -1),"x"))),"")
You will need to expand the range to what you need. Change all the $B$2:$G$7 to $B$2:$N$29. Do not use full column references outside those that I have used. It will kill Excel.
Also note what is and what is not relative references, they need to remain the same or you will get errors as the formula is dragged/copied down.
As simple UDF to do what you want:
Function findMatch(rng As Range, crit As String, inst As Long) As String
Dim rngArr() As Variant
rngArr = rng.Value
Dim i&, j&, k&
k = 0
If k > Application.WorksheetFunction.CountIf(rng, crit) Then
findMatch = ""
Exit Function
End If
For i = LBound(rngArr, 1) + 1 To UBound(rngArr, 1)
For j = LBound(rngArr, 2) + 1 To UBound(rngArr, 2)
If rngArr(i, j) = crit Then
k = k + 1
If k = inst Then
findMatch = rngArr(i, 1) & " " & rngArr(1, j)
Exit Function
End If
End If
Next j
Next i
then you would call it like this:
=findMatch($A$1:$G$7,"x",ROW(1:1))
And drag/copy down.
Related
Does someone know a solution to save the key and the values to an table? My idea does not work because the length of the table is 0 and it should be 3.
local newstr = "3 = Hello, 67 = Hi, 2 = Bye"
a = {}
for k,v in newstr:gmatch "(%d+)%s*=%s*(%a+)" do
--print(k,v)
a[k] = v
end
print(#a)
The output is correct.
run for k,v in pairs(a) do print(k,v) end to check the contents of your table.
The problem is the length operator which by default cannot be used to get the number of elements of any table but a sequence.
Please refer to the Lua manual: https://www.lua.org/manual/5.4/manual.html#3.4.7
When t is a sequence, #t returns its only border, which corresponds to
the intuitive notion of the length of the sequence. When t is not a
sequence, #t can return any of its borders. (The exact one depends on
details of the internal representation of the table, which in turn can
depend on how the table was populated and the memory addresses of its
non-numeric keys.)
Only use the length operator if you know t is a sequence. That's a Lua table with integer indexes 1,..n without any gap.
You don't have a sequence as you're using non-numeric keys only. That's why #a is 0
The only safe way to get the number of elements of any table is to count them.
local count = 0
for i,v in pairs(a) do
count = count + 1
end
You can put #Piglet' code in the metatable of a as method __len that is used for table key counting with length operator #.
local newstr = "3 = Hello, 67 = Hi, 2 = Bye"
local a = setmetatable({},{__len = function(tab)
local count = 0
for i, v in pairs(tab) do
count = count + 1
end
return count
end})
for k,v in newstr:gmatch "(%d+)%s*=%s*(%a+)" do
--print(k,v)
a[k] = v
end
print(#a) -- puts out: 3
The output of #a with method __len even is correct if the table holds only a sequence.
You can check this online in the Lua Sandbox...
...with copy and paste.
Like i do.
I'm trying to iterate through a table with a variable amount of elements and get all possible combinations, only using every element one time. I've landed on the solution below.
arr = {"a","b","c","d","e","f"}
function tablelen(table)
local count = 0
for _ in pairs(table) do
count = count + 1
end
return count
end
function spellsub(table,start,offset)
local str = table[start]
for i = start+offset, (tablelen(table)+1)-(start+offset) do
str = str..","..table[i+1]
end
return str
end
print(spellsub(arr,1,2)) -- Outputs: "a,d,e" correctly
print(spellsub(arr,2,2)) -- Outputs: "b" supposed to be "b,e,f"
I'm still missing some further functions, but I'm getting stuck with my current code. What is it that I'm missing? It prints correctly the first time but not the second?
A solution with a coroutine iterator called recursively:
local wrap, yield = coroutine.wrap, coroutine.yield
-- This function clones the array t and appends the item new to it.
local function append (t, new)
local clone = {}
for _, item in ipairs (t) do
clone [#clone + 1] = item
end
clone [#clone + 1] = new
return clone
end
--[[
Yields combinations of non-repeating items of tbl.
tbl is the source of items,
sub is a combination of items that all yielded combination ought to contain,
min it the minimum key of items that can be added to yielded combinations.
--]]
local function unique_combinations (tbl, sub, min)
sub = sub or {}
min = min or 1
return wrap (function ()
if #sub > 0 then
yield (sub) -- yield short combination.
end
if #sub < #tbl then
for i = min, #tbl do -- iterate over longer combinations.
for combo in unique_combinations (tbl, append (sub, tbl [i]), i + 1) do
yield (combo)
end
end
end
end)
end
for combo in unique_combinations {'a', 'b', 'c', 'd', 'e', 'f'} do
print (table.concat (combo, ', '))
end
For a tables with consecutive integer keys starting at 1 like yours you can simply use the length operator #. Your tablelen function is superfluous.
Using table as a local variable name shadows Lua's table library. I suggest you use tbl or some other name that does not prevent you from using table's methods.
The issue with your code can be solved by printing some values for debugging:
local arr = {"a","b","c","d","e","f"}
function spellsub(tbl,start,offset)
local str = tbl[start]
print("first str:", str)
print(string.format("loop from %d to %d", start+offset, #tbl+1-(start+offset)))
for i = start+offset, (#tbl+1)-(start+offset) do
print(string.format("tbl[%d]: %s", i+1, tbl[i+1]))
str = str..","..tbl[i+1]
end
return str
end
print(spellsub(arr,1,2)) -- Outputs: "a,d,e" correctly
print(spellsub(arr,2,2)) -- Outputs: "b" supposed to be "b,e,f"
prints:
first str: a
loop from 3 to 4
tbl[4]: d
tbl[5]: e
a,d,e
first str: b
loop from 4 to 3
b
As you see your second loop does not ran as the start value is already greater than the limit value. Hence you only print the first value b
I don't understand how your code is related to what you want to achieve so I'll leave it up to you to fix it.
I need to parse numeric range from User Form (VBA, Excel).
For example: {1-3,5,9} -> {1,2,3,5,9,}
The question has already been discussed here:
Advanced parsing of numeric ranges from string
The task is quite standart, is there any ready-made solution for VBA?
I am assuming that the encompassing braces are not part of the string. Consider the following UDF:
Public Function DashFiller(sIn As String) As String
Dim N As Long, NN As Long
DashFiller = ""
ary = Split(sIn, ",")
For N = LBound(ary) To UBound(ary)
If InStr(1, ary(N), "-") > 0 Then
ary2 = Split(ary(N), "-")
ary(N) = ary2(0)
For NN = ary2(0) + 1 To ary2(1)
ary(N) = ary(N) & "," & NN
Next NN
End If
Next N
DashFiller = Join(ary, ",")
End Function
So if A1 contains:
1-2,3,4,5,9-12
and B1 contains:
=DashFiller(A1)
then B1 would display:
1,2,3,4,5,9,10,11,12
I often run into VALUE! errors in my calculations because they contain numbers and text. How can I leave the text in the cell and proceed with the calculation?
For example:
cell A1 contents look like this: 101.1 J
cell A2 contents look like this: 500 U
cell A3 contents look like this: 0.2
If I want to add A1+A2+A3 into cell A4, how can I ignore the J and U to calculate 101.1+500+0.2 to get 602.3 in cell A4?
Thanks!
You need extract the values from the strings - and this can only be done if you have some kind of information about the format to the numbers. In your example, you could place the following formula in B1:B3 and then add a =SUM(B1:B3):
=IF(ISNUMBER(A1),A1,VALUE(LEFT(A1,SEARCH(" ",A1)-1)))
The above formula will extract the number and convert it to a value - unless it was already a number.
Using Custom Function
Place below code in Standard Module
Function add_num(cell1, ParamArray Arr() As Variant)
Dim temp As Double
For i = LBound(Arr) To UBound(Arr)
temp = temp + GetNumber(Arr(i))
Next
add_num = GetNumber(cell1.Value) + temp
End Function
Function GetNumber(ByVal str As String) As Double
Dim objRegEx As Object
Set objRegEx = CreateObject("VBScript.RegExp")
objRegEx.IgnoreCase = True
objRegEx.Global = True
objRegEx.Pattern = "\d{1,2}([\.,][\d{1,2}])?"
Set allMatches = objRegEx.Execute(str)
For i = 0 To allMatches.Count - 1
result = result & allMatches.Item(i)
Next
GetNumber = result
End Function
add_num function can be called from excel interface using =addnum(<cells>). It accepts multiple cells.
How would I go about displaying detailed distance between words.
For example, the output of the program could be:
Words are "car" and "cure":
Replace "a" with "u".
Add "e".
The Levenshtein distance does not fulfill my needs (I think).
Try the following. The algorithm is roughly following Wikipedia (Levenshtein distance). The language used below is ruby
Use as an example, the case of changing s into t as follows:
s = 'Sunday'
t = 'Saturday'
First, s and t are turned into arrays, and an empty string is inserted at the beginning. m will eventually be the matrix used in the argorithm.
s = ['', *s.split('')]
t = ['', *t.split('')]
m = Array.new(s.length){[]}
m here, however, is different from the matrix given if the algorithm in wikipedia for the fact that each cell includes not only the Levenshtein distance, but also the (non-)operation (starting, doing nothing, deletion, insertion, or substitution) that was used to get to that cell from an adjacent (left, up, or upper-left) cell. It may also include a string describing the parameters of the operation. That is, the format of each cell is:
[Levenshtein distance, operation(, string)]
Here is the main routine. It fills in the cells of m following the algorithm:
s.each_with_index{|a, i| t.each_with_index{|b, j|
m[i][j] =
if i.zero?
[j, "started"]
elsif j.zero?
[i, "started"]
elsif a == b
[m[i-1][j-1][0], "did nothing"]
else
del, ins, subs = m[i-1][j][0], m[i][j-1][0], m[i-1][j-1][0]
case [del, ins, subs].min
when del
[del+1, "deleted", "'#{a}' at position #{i-1}"]
when ins
[ins+1, "inserted", "'#{b}' at position #{j-1}"]
when subs
[subs+1, "substituted", "'#{a}' at position #{i-1} with '#{b}'"]
end
end
}}
Now, we set i, j to the bottom-right corner of m and follow the steps backwards as we unshift the contents of the cell into an array called steps, until we reach the start.
i, j = s.length-1, t.length-1
steps = []
loop do
case m[i][j][1]
when "started"
break
when "did nothing", "substituted"
steps.unshift(m[i-=1][j-=1])
when "deleted"
steps.unshift(m[i-=1][j])
when "inserted"
steps.unshift(m[i][j-=1])
end
end
Then we print the operation and the string of each step unless that is a non-operation.
steps.each do |d, op, str=''|
puts "#{op} #{str}" unless op == "did nothing" or op == "started"
end
With this particular example, it will output:
inserted 'a' at position 1
inserted 't' at position 2
substituted 'n' at position 2 with 'r'
class Solution:
def solve(self, text, word0, word1):
word_list = text.split()
ans = len(word_list)
L = None
for R in range(len(word_list)):
if word_list[R] == word0 or word_list[R] == word1:
if L is not None and word_list[R] != word_list[L]:
ans = min(ans, R - L - 1)
L = R
return -1 if ans == len(word_list) else ans
ob = Solution()
text = "cat dog abcd dog cat cat abcd dog wxyz"
word0 = "abcd"
word1 = "wxyz"
print(ob.solve(text, word0, word1))