I'am trying to use this formula to make it work
=ARRAYFORMULA(IF(ISDATE_STRICT(S2:S) ; (MATCH(MAX(AB2:AB),AB2:AB;0)-1) ; "" ))
If there is a date in Column "S" I want it to display the sum of the blanks that would appear if in Column "S" is text
=ARRAYFORMULA(IF(ISDATE_STRICT(S2:S) ; ArrayFormula(MATCH(FALSE ; ISBLANK(AB2:AB) ; 0)-1) ; "" ))
I've tried this one as well but I only get 0's as a result.
Any idea how I can make it work?
Here is the sample sheet.
https://docs.google.com/spreadsheets/d/19f5phXeAwXwrKbWz7njgbznmurOav72GUuo_5IGcbls/edit?usp=sharing
in Q2 use:
=ARRAYFORMULA(IF(ISBLANK(
I1:INDEX(I:I; ROWS(I:I)-1));
{N2:INDEX(N:N; ROWS(N:N))\
I1:INDEX(N:N; ROWS(N:N)-1)};
I1:INDEX(O:O; ROWS(O:O)-1)))
in X2 use:
=INDEX(LAMBDA(x; IFNA(VLOOKUP(x; QUERY(VLOOKUP(ROW(x);
IF(ISDATE_STRICT(x); {ROW(x)\x}); 2; 1);
"select Col1,count(Col1) group by Col1"); 2; 0)-1))
(Q2:INDEX(Q:Q; MAX((Q:Q<>"")*ROW(Q:Q)))))
UPDATE:
we start with column Q. we can take a range Q2:Q but that range contains a lot of empty rows. the next best thing is to check the last non-empty row and set it as the end of the range resulting in Q2:Q73. but static 73 won't do in case the dataset would grow or shrink so to get 73 dynamically we take the MAX of multiplication of Q:Q not being empty and row number of that case eg. Q:Q<>"" will output only TRUE or FALSE so what we are getting is
...
TRUE * 72 = 1 * 72 = 72
TRUE * 73 = 1 * 73 = 73
FALSE * 74 = 0 * 74 = 0
...
so the formula for getting Q2:Q73 is:
=Q2:INDEX(Q:Q; MAX((Q:Q<>"")*ROW(Q:Q)))
it could also be:
=INDEX(INDIRECT("Q2:Q"&MAX((Q:Q<>"")*ROW(Q:Q))))
but it's just long to type... next, we use the new LAMBDA function that allows us to reference cell/range/formula with a placeholder. simple LAMBDA syntax is:
=LAMBDA(x; x)(A1)
where x is A1 and we can do whatever we want with the 2nd (x) argument of LAMBDA like for example:
=LAMBDA(a, a+a*120-a/a)(A1)
you can think of it as:
LAMBDA(A1, A1+A1*120-A1/A1)(A1)
or as just:
=A1+A1*120-A1/A1
the issue here is that we repeat A1 4 times but with LAMBDA we do it only once. also, imagine if we would have 100 characters long formula instead of A1 so the final formula with lambda would be 300 characters shorter compared to "old way" formula.
back to our formula... x is the representation of Q2:Q73. now let's focus on VLOOKUP. basically, the idea here is that IF Q column contains a date we return that date, otherwise we return the last date from above. simply put:
=ARRAYFORMULA(VLOOKUP(ROW(Q2:Q73);
IF(ISDATE_STRICT(Q2:Q73); {ROW(Q2:Q73)\Q2:Q73}); 2; 1))
as you can see Y2, Y3 and Y4 are the same so all we need to do is to count them up and later take away one to exclude Q2 but include just Q3 and Q4 eg. 3-1=2. for that we use simple QUERY where the output is:
date count
30.06.2022 3
so all we need to do is to pair up dates from Q column to QUERY output for that we use the outer VLOOKUP where the output is as follows:
3
#N/A
#N/A
9
#N/A
#N/A
...
now is the right time for that -1 correction while we have these errors coz ERROR-1=ERROR and 3-1=2 so after this -1 correction the output is:
2
#N/A
#N/A
8
#N/A
#N/A
...
and all we need to do now is to hide errors with IFERROR and the output is column X
Related
I'm struggling to understand the behavior of the arguments in the below scan function. I understand the EWMA calc and have made an Excel worksheet to match in an attempt to try to understand but the kdb syntax is throwing me off in terms of what (and when) is x,y and z. I've referenced Q for Mortals, books and https://code.kx.com/q/ref/over/ and I do understand whats going on in the simpler examples provided.
I understand the EWMA formula based on the Excel calc but how is that translated into the function below?
x = constant, y= passed in values (but also appears to be prior result?) and z= (prev period?)
ewma: {{(y*1-x)+(z*x)} [x]\[y]};
ewma [.25; 15 20 25 30 35f]
15 16.25 18.4375 21.32813 24.74609
Rearranging terms makes it easier to read but if I were write this in Excel, I would incorrectly reference the y value column in the addition operator instead of correctly referencing the prev EWMA value.
ewma: {{y+x*z-y} [x]\[y]};
ewma [.25; 15 20 25 30 35f]
15 16.25 18.4375 21.32813 24.74609
EWMA in Excel formula for auditing
0N! is useful in these cases for determining variables passed. Simply add to start of function to display variable in console. EG. to show what z is being passed in as each run:
q)ewma: {{0N!z;(y*1-x)+(z*x)} [x]\[y]};
q)ewma [.25; 15 20 25 30 35f]
15f
16.25
18.4375
21.32812
//Or multiple at once
q)ewma: {{0N!(x;y;z);(y*1-x)+(z*x)} [x]\[y]};
q)
q)ewma [.25; 15 20 25 30 35f]
0.25 15 20
0.25 16.25 25
0.25 18.4375 30
0.25 21.32812 35
Edit:
To think about why z is holding 'y' values it is best to think about below simplified example using just x/y.
//two parameters specified in beginning.
//x initialised as 1 then takes the function result for next run
//y takes value of next value in list
q){0N!(x;y);x+y}\[1;2 3 4]
1 2
3 3
6 4
3 6 10
//in this example only one parameter is passed
//but q takes first value in list as x in this special case
q){0N!(x;y);x+y}\[1 2 3 4]
1 2
3 3
6 4
1 3 6 10
A similar occurrence is happening in your example. x is not being passed to the the iterator and therefore will assume the same value in each run.
The inner function y value will be initilised taking the first value of the outer y variable (15f in this case) like above simplified example. Then the z takes the 2nd value of the list for it's initial run. y then takes the result of previous function run and z takes the next value in the list until how list has bee passed to function.
I am trying to realize my idea in matlab.
I consider two column A and B.
A=data(:,1)
B=data(:,5)
the data look like:
A B
1 1
2 1
3 1
... ...
100 20
... ...
150 30
151 1
... ...
The values in column A are timepoints.
I start with the first element in column A. It schould be A(1,1) and look on the first element in the column B B(1,1). If B(1,1)==1its true,if not its false. Then I increase consider the second raw of the column A and second raw of the column B and so on until the last raw of A and B.
How can I construck this loop??
You can just consider B likes the following:
result = (B == 1);
The result would be the same size of B such as you want. Nowm you can get the value of A on result likes the following:
valid_times = A(result);
I have daily data for a year:
25-Apr-17 45
26-Apr-17 50
27-Apr-17 53
28-Apr-17 47
29-Apr-17 34
30-Apr-17 66
01-May-17 10
02-May-17 42
03-May-17 22
04-May-17 65
05-May-17 76
06-May-17 35
I would like to sum the value, at the month of given date, but prior of the given date ie:
month sum as of date 03-May-17;
I would need to get 10+42+22 = 72 only.
While 04-May-17 value onwards should not be included in sum.
I have tried with sumproduct, sumif but none seems to match this requirement.
Assuming you want this in excel or google sheets.
If you put your dates in col A, values in col B and your criteria date in E1.
Put this formula in another cell (not in col A or B)
=arrayformula(SUM(IF(DAY(A:A)<=DAY($E$1),IF(MONTH(A:A)=MONTH($E$1),B:B,0),0)))
Oddly enough I couldn't do =IF(AND([range of values and compare],[range of values and compare]))
Which would be cleaner but it seemed to evaluate as false
I need to calculate the overall sum without strikeout value
invoice values
--------------------------------
VI 320
VI 260
VI 72
VI 72
-VI- -72-
Just imagine the last value with - symbol is the strikeout value
the total sum should be 724.
i tried doing like the following formula but it return 0 value
IF {Command.void_flg} = 0 THEN
Sum ({Command.Vessel_amt})
First create a new formula #Vessel amt
if ({Command.void_flg}) = 0 THEN
{Command.Vessel_amt}
else 0
Then create another formula #Sum amt.
SUM({#Vessel amt})
Then drag #Sum amtto the place you want.
the thing is that, the 1st number is already ORACLE LONG,
second one a Date (SQL DATE, no timestamp info extra), the last one being a Short value in the range 1000-100'000.
how can I create sort of hash value that will be unique for each combination optimally?
string concatenation and converting to long later:
I don't want this, for example.
Day Month
12 1 --> 121
1 12 --> 121
When you have a few numeric values and need to have a single "unique" (that is, statistically improbable duplicate) value out of them you can usually use a formula like:
h = (a*P1 + b)*P2 + c
where P1 and P2 are either well-chosen numbers (e.g. if you know 'a' is always in the 1-31 range, you can use P1=32) or, when you know nothing particular about the allowable ranges of a,b,c best approach is to have P1 and P2 as big prime numbers (they have the least chance to generate values that collide).
For an optimal solution the math is a bit more complex than that, but using prime numbers you can usually have a decent solution.
For example, Java implementation for .hashCode() for an array (or a String) is something like:
h = 0;
for (int i = 0; i < a.length; ++i)
h = h * 31 + a[i];
Even though personally, I would have chosen a prime bigger than 31 as values inside a String can easily collide, since a delta of 31 places can be quite common, e.g.:
"BB".hashCode() == "Aa".hashCode() == 2122
Your
12 1 --> 121
1 12 --> 121
problem is easily fixed by zero-padding your input numbers to the maximum width expected for each input field.
For example, if the first field can range from 0 to 10000 and the second field can range from 0 to 100, your example becomes:
00012 001 --> 00012001
00001 012 --> 00001012
In python, you can use this:
#pip install pairing
import pairing as pf
n = [12,6,20,19]
print(n)
key = pf.pair(pf.pair(n[0],n[1]),
pf.pair(n[2], n[3]))
print(key)
m = [pf.depair(pf.depair(key)[0]),
pf.depair(pf.depair(key)[1])]
print(m)
Output is:
[12, 6, 20, 19]
477575
[(12, 6), (20, 19)]