Dear Friends,
How to identify the repeat numbers like
111 , Y
122 , N
444 , Y
444.xy , Y # xy is any number
9999 , Y
Maybe converting the int to a string and going through each character of the string to check if they are all same. But i do also think there may be a better approach to this problem
Related
I need to determine whether the following Hash Functions Set is universal or not:
Let U be the set of the keys - {000, 001, 002, 003, ... ,999} - all the numbers between 0 and 999 padded with 0 in the beginning where needed. Let n = 10 and 1 < a < 9 ,an integer between 1 and 9. We denote by ha(x) the rightmost digit of the number a*x.
For example, h2(123) = 6, because, 2 * 123 = 246.
We also denote H = {h1, h2, h3, ... ,h9} as our set of hash functions.
Is H is universal? prove.
I know I need to calculate the probability for collision of 2 different keys and check if it's smaller or equal to 1/n (which is 1/10), so I tried to separate into cases - if a is odd or even, because when a is even the last digit of a*x will be 0/2/4/6/8, else it could be anything. But it didn't help me so much as I'm stuck on it.
Would be very glad for some help here.
If I have a table of prices
t:([]date:2018.01.01+til 30;px:100+sums 30?(-1;1))
date px
2018.01.01 101
2018.01.02 102
2018.01.03 103
2018.01.04 102
2018.01.05 103
2018.01.06 102
2018.01.07 103
...
how do I compute the returns over n days? I am interested in both computing
(px[i] - px[i-n])/px[i-n] and (px[date] - px[date-n])/px[date-n], i.e. one where the column px is shifted n slots by index and one where the previous price is the price at date-n
Thanks for the help
Well you've pretty much got it right with the first one. To get the returns you can use this lambda:
{update return1:(px-px[i-x])%px[i-x] from t}[5]
For the date shift you can use an aj like this:
select date,return2:(px-pr)%pr from aj[`date;t;select date,pr:px from update date:date+5 from t]
Basically what you are trying to do here is to shift the date by the number of days you want to and then extract the price. You use an aj to create your table which will look something like this:
q)aj[`date;t;select date,pr:px from update date:date+5 from t]
date px pr
----------------
2018.01.01 99 98
2018.01.02 98 97
2018.01.03 97 98
Where px is your price now and pr is your price 5 days from now.
Then the return is calculated just the normal way.
Hope this helps!
In tableau, I want to hide alternative rows like:
I have 2 columns name as ID , Name
id name
101 x
102 y
103 z
104 a
Now I want:
id name
101 x
102
103 z
104
I don't believe there is a way to do this simply, as in there is no option dealing with "alternative rows" in Tableau.
Given the sample data provided though, you could create a calculated field to determine the sequence of data you are displaying. For example, the table functions RANK_UNIQUE or INDEX could sequentially assign an integer to your data within the partition. You can then use
[Even_or_odd]:
IF [Sequence_Number] % 2 = 0
THEN "Even"
ELSE "Odd"
You could then use this calculated field to drive a second that would go along the lines of:
[Display_Value]:
IF [Even_or_odd] = Odd
THEN NULL
ELSE [Value]
I have a table field where the data contains our memberID numbers followed by character or character + number strings
For example:
My Data
1234567Z1
2345T10
222222T10Z1
111
111A
Should Become
123456
12345
222222
111
111
I want to get just the member number (as shown in Should Become above). I.E. all the digits that are LEFT of the first character.
As the length of the member number can be different for each person (the first 1 to 7 digit) and the letters used can be different (a to z, 0 to 8 characters long), I don't think I can SPLIT the field.
Right now, in Power Query, I do 27 search and replace commands to clean this data (e.g. find T10 replace with nothing, find T20 replace with nothing, etc)
Can anyone suggest a better way to achieve this?
I did successfully create a formula for this in Excel...but I am now trying to do this in Power Query and I don't know how to convert the formula - nor am I sure this is the most efficient solution.
=iferror(value(left([MEMBERID],7)),
iferror(value(left([MEMBERID],6)),
iferror(value(left([MEMBERID],5)),
iferror(value(left([MEMBERID],4)),
iferror(value(left([MEMBERID],3)),0)
)
)
)
)
Thanks
There are likely several ways to do this. Here's one way:
Create a query Letters:
let
Source = { "a" .. "z" } & { "A" .. "Z" }
in
Source
Create a query GetFirstLetterIndex:
let
Source = (text) => let
// For each letter find out where it shows up in the text. If it doesn't show up, we will have a -1 in the list. Make that positive so that we return the index of the first letter which shows up.
firstLetterIndex = List.Transform(Letters, each let pos = Text.PositionOf(text, _), correctedPos = if pos < 0 then Text.Length(text) else pos in correctedPos),
minimumIndex = List.Min(firstLetterIndex)
in minimumIndex
in
Source
In the table containing your data, add a custom column with this formula:
Text.Range([ColumnWithData], 0, GetFirstLetterIndex([ColumnWithData]))
That formula will take everything from your data text until the first letter.
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)]