So I created a scripting to retrieve data from SAP Gridview object to an Excel sheet. Some columns I needed to replace some characters because this data is consumed by a Power Bi report. For example:
4,350.00 will be replaced for the value 4350. So I do two replaces, the first removing the . and the second replacing the , with .
The problem is that the replace is being applied in every data retrieved. Here's the code.
For i = 0 To GridView.RowCount - 1
For j = 0 To GridView.ColumnCount - 1
shtInput.Cells(z + i, j + 1) = GridView.GetCellValue(i, GridView.ColumnOrder(j))
If j > 8 And j < 19 Then:
rep = GridView.GetCellValue(i, GridView.ColumnOrder(j))
rep = replace(rep, ".", "")
rep = replace(rep, ",", ".")
shtInput.Cells(z + i, j + 1) = rep
Next j
shtInput.Cells(z + i, Area) = "Finance"
If i Mod 32 = 0 Then
GridView.SetCurrentCell i, CStr(Columns(0))
GridView.firstVisibleRow = i
End If
Next i
There's a data column that the script capture the value 21.02.2021 and replace it with 21,02,2021.
Related
Please help as this is getting on my nerves I can't figure out what I'm doing wrong and have tried trace code.
Link to problem: https://leetcode.com/problems/longest-substring-without-repeating-characters/
I created a solution using a sliding window. It works on most test cases, but fails for a few (such as "ad"). I can't figure out where the bug is. I basically keep track in a dictionary of characters I've seen and the last index I saw them at which gets updated periodically in a loop. I use two indices i and j; i gets updated when I find a repeat character. I return the max of current max and length of current substring which is i-j. Here is my code below:
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
if len(s) < 2:
return len(s)
m = 1
i = 0
j = 1
d = {}
d[s[0]] = 0
while j < len(s):
if s[j] in d and d[s[j]] >= i:
m = max(m, j -i)
i = j
d[s[j]] = j
j += 1
return max(m, j - i - 1)
Why does this fail for some cases? Example:
"au"
Output
1
Expected
2
Last line should be return max(m, j - i). Because i is the last index we see repeated character. So. We start this index to end of the string.So length is len(s) - i . And since j = len(s) (while loop ends when j = len(s)) so last substring length is j-i. not j-i-1
And also we are updating i wrong.let's say s = "abcadf". In while loop when we see second "a" ,so j = 3, we should update i = 1, not 3. Because in this case our longest substring will start with "b".So we should update i as i = d[s[j]] + 1. So final result:
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
if len(s) < 2:
return len(s)
m = 1
i = 0
j = 1
d = {}
d[s[0]] = 0
while j < len(s):
if s[j] in d and d[s[j]] >= i:
m = max(m, j -i)
i = d[s[j]] + 1
d[s[j]] = j
j += 1
return max(m, j - i)
Help me I need to create a Userform for my personal calculations. But I ran into a lot of problems. Because I've never written a program before.
When I enter a value 0 or enter a value other than 0 and delete it in the text field PriceCoinBuy, BuyCoin , PriceCoinSell , SellCoin , Vat one of the fields I will get Msg : Run-time error '6'; overflow.
But when I put a number other than 0 in the BuyCoin field, I get Msg : Run-time error '11'; Division by zero.
I've been searching for a solution for 2 days but can't find it in please help.
I have 5 textboxes for input.
Sub SumAll()
Dim A, B, C, D, E, F, G, H, I, J, K, V As Double
A = Val(Order.PriceCoinBuy.Value)
B = Val(Order.BuyCoin.Value)
C = Val(Order.PriceCoinSell.Value)
D = Val(Order.SellCoin.Value)
V = Val(Order.Vat.Value)
'-------------- Math --------------
E = CDbl(B) / A
F = CDbl(E) * (V / 100)
G = CDbl(E) - F
H = CDbl(G) * A
I = CDbl(D) * C
J = CDbl(I) * (V / 100)
K = CDbl(I) - J
'---------------- Show -------------
Order.GetCoin.Text = Format(E, "##,##0.000000")
Order.AfterVatBuy.Text = Format(F, "##,##0.0000")
Order.CoinBalance.Text = Format(G, "##,##0.000000")
Order.ToMoney.Text = Format(H, "##,##0.00")
Order.GetMoney.Text = Format(I, "##,##0.00")
Order.AfterVatSell.Text = Format(J, "##,##0.00")
Order.MoneyBalance.Text = Format(K, "##,##0.00")
End Sub
I am trying to figure out a way to make a for loop in which I can compare two cells that will give me two different means. One for class char and the other for class double.
This is what I have so far.
V = {2; 'tree'; 3; 'hope'};
W = {2; 'tree'; 3; 'hope'};
for i = 1:length(V);
if isequal(class(V{i}), 'double')
num = V{i}
elseif isequal(class(V{i}), 'char')
str = V{i}
end
end
for i = 1:length(W);
if isequal(class(W{i}), 'double')
acc_n(i) = isequal(V{i}, W{i})
elseif isequal(class(W{i}), 'char')
acc_s(i) = strcmp(V{i}, W{i})
end
end
mean_d = mean(acc_n)
mean_s = mean(acc_s)
The output I get is:
acc_n =
1 0 1
acc_s =
0 1 0 1
mean_d =
0.6667
mean_s =
0.5000
The output I want is:
1 1 for string, mean = 1. 1 1 for double, mean = 1
How can I do a loop where it only takes the numbers of the cell and the words of the cell separately?
Is there any possible way to only loop through the words or the numbers?
You can first extract strings and doubles and treat them separately, that will avoid loops.
V = {2; 'tree'; 3; 'hope'};
W = {2; 'tree'; 3; 'hope'};
VChar=V(cellfun(#ischar,V));
WChar=W(cellfun(#ischar,W));
acc_s=VChar==WChar;
VNum=cell2mat(V(cellfun(#isnumeric,V)));
WNum=cell2mat(W(cellfun(#isnumeric,W)));
acc_n=VNum==WNum;
Loop version: I haven't tested this but it should work.
%Assumes V and W have equal number of elements.
acc_n=[];
acc_s=[];
for i=1:numel(V)
if isequal(class(V{i}), 'double') && isequal(V{i},W{i})
acc_n=[acc_n true];
elseif isequal(class(V{i}), 'char') && strcmp(V{i},W{i})
acc_s=[acc_s true];
end
end
I have a group of people who have different expenses during a period. Everybody has a balance after this period. It looks like this in excel:
Person A: 6
Person B: 10
Person C: -7,5
Person D: -8,5
After this period a settlement will take place. I currently do this by manually. This results in:
Person C pays 6 to person A.
Person C pays 1,5 to person B.
Person D pays 8,5 to person B.
Person A gets 6 from person C.
Person B gets 1,5 form person C.
Peron B gets 8,5 from person D.
(There are multiple solutions possible when more persons are involved.)
The problem is that I have to apply this procedure for a big group of people. So my question is: 'How to apply this 'who owes who'-procedure by using an excel spreadsheet algorithm or macro?'.
I made a excel version too but its in Open office. can you download that? The following macro might work on its own. If it does not should be something small. it works fine in OOO and is saved as a Excel 97/2000 workbook document.
'this might not be needed in Microsoft Excel, comment it out
Option VBASupport 1 'OWES
Option Explicit
'Cells(row, col),
Private Sub cmd1_Click()
'data is same sheet, from row 4, column 4
'row 4 has names in columns, so 4,4 has name 1, 4,5 has name 2
'row 5 has amounts spent like 6, -10
'output is in columns 3 and 5
dim i
dim j,s as String, sum1
s=""
'get number of cells used in row 4 and check if corresponding row 6 column has a number value too
i = 4
sum1=0
do while(cells(4,i).Value <> "" and i < 500)
j = CDbl(cells(5,i).Value)
sum1 = sum1 + j
if j <> cells(5,i).Value then
MsgBox "Col " & i & " not a number?"
End
end if
i=i+1
loop
if i > 499 then
Msgbox "too many cols"
End
end if
If sum1 > 0.3 or sum1 < -0.3 then
Msgbox "Sum is not near 0 :" & sum1
End
End if
Dim colCnt as Integer
colCnt = i - 4
cells (7,1).Value = "Col count = " & colCnt
Dim spent(colCnt) as Double
Dim owes1(colCnt ) as String
Dim owes2(colCnt ) as String
for i= 4 to colCnt + 3
spent(i - 3) = CDbl(cells(5,i).Value)
Next
Dim cnt,lastNeg, abs1,maxPay ' safety var for never ending loops, only if data bad like many cols and more than .1 diffs
lastNeg = 4
dim lastPay1
lastPay1 = 10
dim ii,jj,c1,c2,toPay
toPay = 0
On Local Error Goto errh
for i= 4 to colCnt + 3
cnt = 0
ii = i - 3
c1 = spent(ii)
'Cells(6,i) = "ok "
if spent(ii) > 0.1 and cnt < colCnt Then '//has to take
cnt = cnt + 1
for j = lastNeg to colCnt + 3 ' ; j < people.length && spent(ii) > 0.1; j++)
jj = j - 3
's = s & Me.Cells(ii,j) & " "
if spent(ii) > 0.1 then
if spent(jj) < -0.1 Then ' //has to give and has balance to give
c1 = spent(ii)
c2 = spent(jj)
lastNeg = j
abs1 = spent(jj) * -1'//can use absolute fn
maxPay = abs1
if(maxPay > spent(ii)) Then
toPay = spent(ii)'
else
toPay = abs1
End if
spent(ii) = spent(ii) - toPay
spent(jj) = spent(jj) + toPay
Cells(lastPay1, 3).Value = Cells(4 , j) & " pays " & toPay & " to " & Cells(4 , i )
Cells(lastPay1, 5).Value = Cells(4 , i) & " gets " & toPay & " from " & Cells(4 , j)
lastPay1 = lastPay1 + 1
End if
End if
Next
End if
Next
Msgbox "Done"
err.Clear
if err.Number <> 0 Then
errH:
dim yy
yy = msgBox("err " & err.Number & " " & err.Description & " Continue", 2)
if yy = vbYes Then
Resume Next
End IF
End IF
End Sub
Book at http://sel2in.com/prjs/vba/profile (owes)
Can see http://www.excel-vba.com/ , http://office.microsoft.com/en-in/training/get-in-the-loop-with-excel-macros-RZ001150634.aspx the help in excel was useful too (f1 inside macro editor, can select a keyword or type and get context sensitive help by pressing f1)
How important is the pairing of who owes what to who? The reason I ask - it's simple to figure out the total cost per person and then determine who owes money and who needs a refund. If one person can be the "banker", he can collect all the money due and disburse all the refunds.
Much simpler question, if you have somebody willing to be the banker.
Trying to pair everything up will quite possibly not result in an exact answer, or may require one or more people making payments to more than one person - as well as one or more people trying to collect from more than one other person.
See http://sel2in.com/pages/prog/html/owes.html
Javascript fn
<form>
Paste tab seperated list of people on first line, Second line has blanace - positive means they spent more than others, negative means other spent on them (they owe)
<br>
<textarea id=t1 rows=3 cols=70></textarea>
</form>
<button onclick=calcOwes() >Calc owes</button>
<br>
<b>Result:</b>
<div id=result>Will appear here if data is good</div>
<br>
<b>Parsed data for debug:</b>
<div id=data1>Will appear here if data is good</div>
<br>
<hr>
<script>
function calcOwes(){
/**
2013 Tushar Kapila
If Person A: 6
Person B: 10
Person C: -7,5
Person D: -8,5
Then Person C pays 6 to person A.
Person C pays 1,5 to person B.
Person D pays 8,5 to person B.
*/
s = document.getElementById("t1").value
var line = s.split("\n")
//v = confirm("Your Data looks okay?:\n" + s)
//if(!v){ return;}
if(s.length < 2 || s.indexOf("\n") < 0){
alert("No line sep ")
return
}
people = line[0].split("\t")
spent = line[1].split("\t")
spent2 = line[1].split("\t")
if(spent.length < 2){
alert("Bad data, no spent data " + spent.length + "; 0 " + spent[0] + "; 1 " + + spent[1])
return
}
if(people.length != spent.length){
alert("People and amounts do not tally. make sure no tabs inside names, spaces are okay")
return
}
sum = 0;
data1o = document.getElementById("data1")
data1o.innerHTML = "";
for(i = 0;i < people.length; i++){
spent[i] = spent[i].trim()
spent[i] = parseFloat(spent[i])
sum += spent[i]
s = (1 + i) + " \"" + people[i] + "\" :" + spent[i] + ";<br>"
data1o.innerHTML += s;
}
if(sum > 0.2 || sum < -0.2){
v = confirm("Sum (" + sum + ")is not zero continue?")
if(!v){return;}
}
lastNeg = 0;
payDetails = new Array();
getDetails = new Array();
lastPay = 0;
for(i = 0;i < people.length; i++){
cnt = 0;
if(spent[i] > 0.1 && cnt < people.length){//has to take
cnt++
for(j = lastNeg; j < people.length && spent[i] > 0.1; j++){
if(spent[j] < -0.1){//has to give and has balance to give
lastNeg = j;
abs1 = spent[j] * -1;//can use absolute fn
maxPay = abs1
if(maxPay > spent[i]){
toPay = spent[i];
}else{
toPay = abs1
}
spent[i] -= toPay
spent[j] += toPay
payDetails[lastPay] = people[j] + " pays " + toPay + " to " + people[i]
getDetails[lastPay] = people[i] + " gets " + toPay + " from " + people[j]
lastPay++;
}
}
}
}
s = ""
s2 = ""
for(i = 0;i < lastPay; i++){
s = s + payDetails[i] + "<br>"
s2 = s2 + getDetails[i] + "<br>"
}
document.getElementById("result").innerHTML = s + "<br>" + s2
}
</script>
<br>Sample input 1 (tabs there?)
<br><pre>
a b c d
6 10 -7.5 -8.5
</pre>
<br>Sample input 2
<pre>
Anna Dan Bobby Scareface Colly Doc Egg face
-6 10 -7.3 -8.33 11.67
</pre>
I am new to VBA coding. I have done some coding in Javascript and C++, so I do understand the concepts. I'm not too familiar with the specifics of VBA, though. This particular code is for Excel 2007. The sort function was copied from elsewhere as pseudocode (documentation is not mine). I've rewritten it as VBA (unsuccessfully).
This code is not working properly. The code is abruptly aborting entirely (not just jumping out of a loop or function, but quitting completely after going through the While loop twice.
To replicate the problem, save this code as a Macro for an Excel sheet, type the number 9853 in B5, and in B6 type "=Kaprekar(B5)". Essentially, run Kaprekar(9853).
Could someone please help me figure out what I'm doing wrong here? Thanks.
By the way, I'm using While-Wend now. I also tried Do While-Loop with the same result.
Here's the code:
Function Sort(A)
limit = UBound(A)
For i = 1 To limit
' A[ i ] is added in the sorted sequence A[0, .. i-1]
' save A[i] to make a hole at index iHole
Item = A(i)
iHole = i
' keep moving the hole to next smaller index until A[iHole - 1] is <= item
While ((iHole > 0) And (A(iHole - 1) > Item))
' move hole to next smaller index
A(iHole) = A(iHole - 1)
iHole = iHole - 1
Wend
' put item in the hole
A(iHole) = Item
Next i
Sort = A
End Function
Function Kaprekar%(Original%)
Dim Ord(0 To 3) As Integer
Ord(0) = Original \ 1000
Ord(1) = (Original - (Ord(0) * 1000)) \ 100
Ord(2) = (Original - (Ord(1) * 100) - (Ord(0) * 1000)) \ 10
Ord(3) = (Original - (Ord(2) * 10) - (Ord(1) * 100) - (Ord(0) * 1000))
If (Ord(0) = Ord(1)) * (Ord(1) = Ord(2)) * (Ord(2) = Ord(3)) * (Ord(3) = Ord(0)) = 1 Then
Kaprekar = -1
Exit Function
End If
Arr = Sort(Ord)
Kaprekar = Ord(3)
End Function
excel evaluates both items in the while statement, so
While ((ihole > 0) And (A(ihole - 1) > item))
when ihole=0, returns false for the first test, and out of bounds for the second test, bombing out of the function with a #Value error.
A quick bubblesort would be something like this:
Option Explicit
Function Sort(A)
Dim iLoop As Long
Dim jLoop As Long
Dim Last As Long
Dim Temp
Last = UBound(A)
For iLoop = 0 To Last - 1
For jLoop = iLoop + 1 To Last
If A(iLoop) > A(jLoop) Then
Temp = A(jLoop)
A(jLoop) = A(iLoop)
A(iLoop) = Temp
End If
Next jLoop
Next iLoop
Sort = A
End Function