I am currently working on a traveling sales person problem in Excel. For this I need to generate two rows parallel to each other ( Row1 for row Row2 for column) and use INDEX MATCH MATCH to look up the distances between nodes and solve using solver in order to get the best sequence to minimize distance.
For example I have row0 of four nodes named [1][2][3][4].
Distance matrix
__[1][2][3][4]
[1] 0, 4, 5, 7
[2] 4, 0, 6, 2
[3] 5, 6, 0, 1
[4] 7, 2, 1, 0
It must take row 0 and using a macro to develop the following 2 rows
Row 1 [1][1][1][2][2][3]
Row 2 [2][3][4][3][4][4]
Then I can use the Index Match Match to automatically bring all the distances below in a third row:
Row 1 [1][1][1][2][2][3]
Row 2 [2][3][4][3][4][4]
Row 3 4, 5, 7, 6, 2, 1
Please I need help developing the macro to create row 1 an row 2. This should be able to account for any amount of nodes and not just the 4 listed in the example.
I got the answer for row 2. Sorry if it seems like a basic question but I am a noob at VBA and hoping to get better. Ill post when I have the answer for the next row.
Sub Macro1()
'
' Macro1 Macro
k = 5
i = 0
j = 0
rwar = Worksheets("Sheet1").Cells(13, 5).Value
For i = 3 To rwar + 1
rwar = rwar - 1
For j = 1 To rwar
Worksheets("Sheet1").Cells(16, k) = Worksheets("Sheet1").Cells(i, 5).Value
k = k + 1
Next j
Next i
'
End Sub
Related
Fellow overflowers, all help is appreciated;
I have the following rows of values (always 7 values per row) of data in Excel (3 examples below), where data is coded as 1 or 2. I am interested in the 1's.
2, 2, 1, 2, 2, 1, 1.
1, 2, 2, 2, 2, 1, 2.
2, 2, 2, 1, 1, 1, 2.
I use the =MATCH(1,A1:G1,0) to tell me WHEN the first 1 appears, BUT now I want to calculate the proportion that 1's make up of the the remaining values in the row.
For example;
2, 2, 1, 2, 2, 1, 1. (1 first appears at point 3, but then 1's make up 2 out of 4 remaining points; 50%).
1, 2, 2, 2, 2, 1, 2. (1 first appears at point 1, but then 1's make up 1 out of the 6 remaining points; 16%).
2, 2, 2, 1, 1, 1, 2. (1 first appears at point 4, but then 1's make up 2 out of the 3 remaining points; 66%).
Please help me calculate this proportion!
You could use this one
=(LEN(SUBSTITUTE(SUBSTITUTE(MID(A1,SEARCH(1,A1)+3,1000)," ",""),",",""))
-LEN(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(MID(A1,SEARCH(1,A1)+3,1000)," ",""),",",""),1,""))
)/LEN(SUBSTITUTE(SUBSTITUTE(MID(A1,SEARCH(1,A1)+3,1000)," ",""),",",""))
The
SUBSTITUTE(SUBSTITUTE(MID(A1,SEARCH(1,A1)+3,1000)," ",""),",","")
-part gets the string after the first 1. The single 1 in the middle part is the one, you want to calculate the percentage for. So if you want to adapt the formula to other chars, you have to change the single 1 in th emiddle part and the three 1s in the three searches.
EDIT thank you for the hint #foxfire
A solution for values in columns would be
=COUNTIF(INDEX(A1:G1,1,MATCH(1,A1:G1,0)+1):G1,1)/(COUNT(A1:G1)-MATCH(1,A1:G1,0))
You can do it with SUMPRODUCT:
My formula in column H is a MATCH like yours:
=MATCH(1;A3:G3;0)
My formula for calculatin % of 1's over reamining numbers after first 1 found, is:
=SUMPRODUCT((A3:G3=1)*(COLUMN(A3:G3)>H3))/(7-H3)
This is how it works:
(A3:G3=1) will return an array of 1 and 0 if cell value is 1 or not. So for row 3 it would be {0;0;1;0;0;1;1}.
COLUMN(A3:G3)>H3 will return an array of 1 and 0 if column number of cell is higher than column number of first 1 found, (that matchs with its position inside array). So for row 3 it would be {0;0;0;1;1;1;1}
We multiply both arrays. So for row 3 it would be {0;0;1;0;0;1;1} * {0;0;0;1;1;1;1} = {0;0;0;0;0;1;1}
With SUMPRODUCT we sum up the array of 1 and 0 from previous step. So for row 3 we would obtain 2. That means there are 2 cells with value 1 after first 1 found.
(7-H3) will just return how many cells are after first 1 found, so fo row 3, it means there are 4 cells after first 1 found.
We divide value from step 4 by value from previous step, and that's the % you want. So for row 3, it would be 2/4=0,50. That means 50%
update: I used 2 columns just in case you need to show where is the first 1. But in case you want a single column with the %, formula would be:
=SUMPRODUCT((A3:G3=1)*(COLUMN(A3:G3)>MATCH(1;A3:G3;0)))/(7-MATCH(1;A3:G3;0))
I have 4 columns of data - B10:E(LastRow).
Coulmn 1 = Dates - Not needed
Coulmn 2 = Money - Not Needed
Coulmn 3 = Segments Sold - Needed
Column 4 = Segments Left - Needed
I am struggling to write the code for Column 4.
I have a fixed value of segments called from Cells(4, 3), which is 100 in this case, to start the if statement.
I know my below code is incorrect.
For k = 10 To LR1
If ws2.Cells(k, "D") = 0 Then
ws2.Cells(k, "E") = 100
ElseIf ws2.Cells(k, "D") > 0 Then
ws2.Cells(k, "E") = ws2.Cells(4, 3) - ws2.Cells(k, "D")
End If
Next k
Basically I want Cells 10: to the first value that is > 0 to equal 100.
Then I want the cells 11th, from the first value greater than 0, we shall say 3, to be 97, until the next value greater than 0, say 7, thus it will equals 90 and so on for the rest of the column.
I hope this is clear enough. I have attached an image, left one is what I currently get, right one is what I would like to achieve.
Left - What I get with the if function // Right - What I want to achieve
I have the following problem: I have a matrix opened with pandas module, where each cell has a number between -1 and 1. What I wanted to find is the maximum "posible" value in a row that is also not the maximum value in another row.
If for example 2 rows has their maximum value at the same column, I compare both values and take the bigger one, then for the row that has its maximum value smaller that the other row, I took the second maximum value (and do the same analysis again and again).
To explain myself better consider my code
import pandas as pd
matrix = pd.read_csv("matrix.csv")
# this matrix has an id (or name) for each column
# ... and the firt column has the id of each row
results = pd.DataFrame(np.empty((len(matrix),3),dtype=pd.Timestamp),columns=['id1','id2','max_pos'])
l = len(matrix.col[[0]]) # number of columns
while next = 1:
next = 0
for i in range(0, len(matrix)):
max_column = str(0)
for j in range(1, l): # 1 because the first column is an id
if matrix[max_column][i] < matrix[str(j)][i]:
max_column = str(j)
results['id1'][i] = str(i) # I coul put here also matrix['0'][i]
results['id2'][i] = max_column
results['max_pos'][i] = matrix[max_column][i]
for i in range(0, len(results)): #now I will check if two or more rows have the same max column
for ii in range(0, len(results)):
# if two id1 has their max in the same column, I keep it with the biggest
# ... max value and chage the other to "-1" to iterate again
if (results['id2'][i] == results['id2'][ii]) and (results['max_pos'][i] < results['max_pos'][ii]):
matrix[results['id2'][i]][i] = -1
next = 1
Putting an example:
#consider
pd.DataFrame({'a':[1, 2, 5, 0], 'b':[4, 5, 1, 0], 'c':[3, 3, 4, 2], 'd':[1, 0, 0, 1]})
a b c d
0 1 4 3 1
1 2 5 3 0
2 5 1 4 0
3 0 0 2 1
#at the first iterarion I will have the following result
0 b 4 # this means that the row 0 has its maximum at column 'b' and its value is 4
1 b 5
2 a 5
3 c 2
#the problem is that column b is the maximum of row 0 and 1, but I know that the maximum of row 1 is bigger than row 0, so I take the second maximum of row 0, then:
0 c 3
1 b 5
2 a 5
3 c 2
#now I solved the problem for row 0 and 1, but I have that the column c is the maximum of row 0 and 3, so I compare them and take the second maximum in row 3
0 c 3
1 b 5
2 a 5
3 d 1
#now I'm done. In the case that two rows have the same column as maximum and also the same number, nothing happens and I keep with that values.
#what if the matrix would be
pd.DataFrame({'a':[1, 2, 5, 0], 'b':[5, 5, 1, 0], 'c':[3, 3, 4, 2], 'd':[1, 0, 0, 1]})
a b c d
0 1 5 3 1
1 2 5 3 0
2 5 1 4 0
3 0 0 2 1
#then, at the first itetarion the result will be:
0 b 5
1 b 5
2 a 5
3 c 2
#then, given that the max value of row 0 and 1 is at the same column, I should compare the maximum values
# ... but in this case the values are the same (both are 5), this would be the end of iterating
# ... because I can't choose between row 0 and 1 and the other rows have their maximum at different columns...
This code works perfect to me if I have a matrix of 100x100 for example. But, if the matrix size goes to 50,000x50,000 the code takes to much time in finish it. I now that my code could be the most inneficient way to do it, but I don't know how to deal with this.
I have been reading about threads in python that could help but it doesn't help if I put 50,000 threads because my computer doesn't use more CPU. I also tried to use some functions as .max() but I'm not able to get column of the max an compare it with the other max ...
If anyone could help me of give me a piece of advice to make this more efficient I would be very grateful.
Going to need more information on this. What are you trying to accomplish here?
This will help you get some of the way, but in order to fully achieve what you're doing I need more context.
We'll import numpy, random, and Counter from collections:
import numpy as np
import random
from collections import Counter
We'll create a random 50k x 50k matrix of numbers between -10M and +10M
mat = np.random.randint(-10000000,10000000,(50000,50000))
Now to get the maximums for each row we can just do the following list comprehension:
maximums = [max(mat[x,:]) for x in range(len(mat))]
Now we want to find out which ones are not maximums in any other rows. We can use Counter on our maximums list to find out how many of each there are. Counter returns a counter object that is like a dictionary with the maximum as the key, and the # of times it appears as the value.
We then do dictionary comprehension where the value is == to 1. That will give us the maximums that only show up once. we use the .keys() function to grab the numbers themselves, and then turn it into a list.
c = Counter(maximums)
{9999117: 15,
9998584: 2,
9998352: 2,
9999226: 22,
9999697: 59,
9999534: 32,
9998775: 8,
9999288: 18,
9998956: 9,
9998119: 1,
...}
k = list( {x: c[x] for x in c if c[x] == 1}.keys() )
[9998253,
9998139,
9998091,
9997788,
9998166,
9998552,
9997711,
9998230,
9998000,
...]
Lastly we can do the following list comprehension to iterate through the original maximums list to get the indicies of where these rows are.
indices = [i for i, x in enumerate(maximums) if x in k]
Depending on what else you're looking to do we can go from here.
Its not the speediest program but finding the maximums, the counter, and the indicies takes 182 seconds on a 50,000 by 50,000 matrix that is already loaded.
How can I generate random numbers 0 or 1 in 10 cells in the row, in which the sum of the random number is always equal to 7?
enter image description here
Here's a way to get seven "1"s and three "0"s in random order using RAND and RANK
In A1:J1: =RAND()
In A2:J2: =IF(RANK(A1,$A$1:$J$1,1)>3,1,0)
Available here is a version that I really think works! https://www.dropbox.com/s/ec431fu0h0fhb5i/RandomNumbers.xlsx?dl=0
And here's the '0 and 1' version (sheet 2 at the above link):
De-dup Rankings Randoms First Cut Sorted
0.47999002 7 0.479992063 1 1
0.68823003 3 0.688233075 1 1
0.07594004 9 0.075938331 1 1
0.02077005 10 0.020766892 1 0
0.69217006 2 0.692170173 1 0
0.73355007 1 0.733549516 1 1
0.51546008 6 0.515462872 1 1
0.62308009 4 0.623078278 0
0.33033001 8 0.330331577 1
0.561260011 5 0.561260557 1
Formulae for columns A-C exactly as before, D is just 7 1's, E is:
=VLOOKUP(ROW(E2)-1,B$1:D$11,3,FALSE)
Assuming that you want a list of positive random numbers that add to 7 you can use this following method.
Enter a 0 in the top-left cell (Blue Cell).
Enter =RAND()*7 into the next 9 cells below the 0 (Orange Cells).
Enter a 7 in the cell below the 9 random values (Blue Cell).
Copy the 9 random values and paste-special-values over top to turn the formulas into values.
Sort just these 9 cells in ascending order
In the cell just to the right of the first random value put a formula that subtracts the cell to the left and one above from the cell to the left (Yellow Cells).
Repeat this formula down to the cell next to the 7 that was typed in.
Sum the values in the second column (Green Cell).
That should give you 10 random values whose sum is exactly 7.
The only issue is that getting the values to be between 0 and 1 will take a bit of trial and error.
It appears that trial and error may not be practical. It's about a one in 2,710 times that this list will contain only numbers between 0 and 1. Not overly practical. Sorry.
To answer the question in the post, enter this in A1:J1 as an array formula (ctrl+shift+enter):
=1-(TRANSPOSE(MOD(SMALL(RANDBETWEEN(0,1e12*(ROW(INDIRECT("1:10"))>0))+(ROW(INDIRECT("1:10"))-1)/10,ROW(INDIRECT("1:10"))),1))>0.65)
To answer the question in the post title, do the following:
In A1:J1 enter:
=RAND()
In K1 enter:
=IF(SUM(A1:J1)<7,(7-SUM(A1:J1))/(COUNT(A1:J1)-7),7/SUM(A1:J1))
In L1 enter:
=IF(SUM($A1:$J1)<7,(A1+$K1)/($K1+1),A1*$K1)
Fill over to U1.
I believe the 10 numbers generated will be identically distributed in [0,1), but obviously not uniformly (I'm fairly certain the distribution does not have a name). The numbers can't be considered independent. A few statistics on the distribution:
Mean: 0.7 (as expected)
The other statistics are estimated from 10,000 simulations:
Variance: 0.0295
Kurtosis: -0.648
Skewness: -0.192
Think of it as drawing a sample of size 7 from the set {1, 2, ..., 10}. The 1s correspond to the numbers chosen for inclusion in the sample. Here is some VBA code which generates such samples:
Function sample(n As Long, k As Long) As Variant
'returns a variant of length n
'consisting of k 1s and n-k 0s
'thought of as a sample of {1,...,n} of size k
Dim v As Variant 'vector to hold sample
Dim numbers As Variant
Dim i As Long, j As Long, temp As Long
ReDim v(1 To n)
ReDim numbers(1 To n)
For i = 1 To n
v(i) = 0
numbers(i) = i
Next i
'do k steps of a Fisher-Yates shuffle on numbers
For i = 1 To Application.WorksheetFunction.Min(k, n - 1)
j = Application.WorksheetFunction.RandBetween(i, n)
If i < j Then 'swap
temp = numbers(i)
numbers(i) = numbers(j)
numbers(j) = temp
End If
Next i
'now use the first k elements of the partially shuffled array:
For i = 1 To k
v(numbers(i)) = 1
Next i
sample = v
End Function
Used like: Range("A1:J1").Value = sample(10,7)
Using a bit of brute force, I think I've got a workable solution to the original version of the question which asked for random numbers between 0 and 1.
Cells A1 to A9:
=rand()
Cell A10:
=7-sum(A1:A9)
Now you have 10 numbers that add up to 7, but the last one is probably not in the range 0 to 1. To deal with that, just recalculate the sheet to generate new random numbers until that last value is within range. It takes about 25 recalculations to have a ~95% chance that one of them will be within range, so it could take a while. A little VBA can do that for you very quickly:
Sub rand7()
While Range("A10").Value > 1 Or Range("A10").Value < 0
ActiveSheet.Calculate
Wend
End Sub
I need a macro or formula that can do this:
Column A Values:
1
1
1
2
2
2
3
3
4
4
4
5
5
6
6
I need Column B to do:
1
2
3
1
2
3
1
2
1
2
3
1
2
1
2
Because column B has to have an increment number while column A has the same repeated value (for example 1 1 1 2), when it changes (to 2 or 3 etc.) the counter on column B has to reset and increment itself while A repeats the next value (1 2 3 1)
Thanks
Edit
Have a look at bzimor's answer as he solves is in a more general way without a hard value in column B1!
Original Answer
You can solve it the following way without VBA:
Put a hard value of 1 in column B1. This is your start value and should be always correct because you start counting with 1.
Then enter the following formula in B2
=IF($A2=$A1;$B1+1;1)
Just drag the formula down to the other rows of B and you're done
So B3 should look like this
=IF($A3=$A2;$B2+1;1)
and so on ...
You can use single formula:
=COUNTIF($A$1:A1,A1)
Put it into cell B1 and fill down
Here is your desired macros
Sub jkjainGenerateSerialNumber20161124()
readcol = Val(InputBox("type ref col#"))
writecol = Val(InputBox("type dest col#"))
ts = 0
Range("A1000000").Select
Selection.End(xlUp).Select
lastrow = ActiveCell.Row
For n = 2 To lastrow
prev = Cells(n - 1, readcol)
curr = Cells(n, readcol)
If prev <> curr Then
ts = 1
End If
If prev = curr Then
ts = ts + 1
End If
Cells(n, writecol) = ts
Next
End Sub