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.
Related
I have a table of numbers that are in an array that have gotten mapped and now I'm trying to present them right aligned for example I have this:
[1,2,3,4,5,6]
[1,2,44,5,66,77]
But want this:
1 2 3 4 5 6
1 2 44 5 66 77
Not sure if its coming through but I don't want the brackets or quotes if the values were a string BUT I want them right aligned vs left aligned. I figured out left aligned and just trying to see if there is an easy way to do this.
var arr= [0,1,2,3]
for i in 0...3 {
let table = arr.map { $0 * i }
print (table)
}
You are simply printing the array and the description method of Array will show the list of values separated by commas with the brackets.
If you want any other output you need to generate it yourself.
Replace your current print with the following:
let line = table.map { String(format: "%4d", $0)}.joined()
print(line)
This maps the array of Int into an array of String and then joins those strings into a single string with no separator between them. Each Int is formatted into a String that will take four spaces and the number will be right-aligned within those four spaces. Adjust as needed.
The current raw data :
1-2-05.11
1-15-05.20
how can I remove after .. The expected result is 1-2-05. I test using split_part and also substring, but the result is not fit the requirement.
Any suggestion ?
Try this,I assume that your expected output is 1-2-05.(with .)
Using split_part().
SELECT SPLIT_PART('1-2-05.11','.',1)||'.';
Using substring().
SELECT SUBSTRING('1-15-05.20', 1, LENGTH('1-15-05.20') - 2)
1 is the starting position(from left) of the string(1-15-05.20) in which substring action to be taken
LENGTH('1-15-05.20') - 2, is to define the number of character to be extracted from the given string.The string is 1-15-05.20 the length() of it is 10, you need to remove last two characters from this 10 chars so 10 - 2 ie LENGTH('1-15-05.20') - 2
I am new in Swift.
I am trying to make a budget application. This app have a Calculator like keyboard. My idea is when users enter the money app will automatically add a decimal place for users.
For example, if you type 1230 it will give you 12.30 and type 123 it will display 1.23
I wrote a couple lines of code down below. The problem is it only can add decimal point after first digit it won't go backwards when you give more digits. It only can display as X.XXXXX
I tried solve this problem with String.index(maybe increase index?) and NSNumber/NSString format. But I don't know this is the right direction or not.
let number = sender.currentTitle!
let i: String = displayPayment.text!
if (displayPayment.text?.contains("."))!{
displayPayment.text = i == "0" ? number : displayPayment.text! + number
}
else {
displayPayment.text = i == "0" ? number : displayPayment.text! + "." + number
}
Indexing Strings in Swift is not as "straightforward" as many would like, simply due to how Strings are represented internally. If you just want to add a . at before the second to last position of the user input you could do it like this:
let amount = "1230"
var result = amount
if amount.characters.count >= 2 {
let index = amount.index(amount.endIndex, offsetBy: -2)
result = amount[amount.startIndex..<index] + "." + amount[index..<amount.endIndex]
} else {
result = "0.0\(amount)"
}
So for the input of 1230 result will be 12.30. Now You might want to adjust this depending on your specific needs. For example, if the user inputs 30 this code would result in .30 (this might or might not be what you want).
I have already used the text delimiters and item numbers to extract a date from a file name, so I'm clear about how to use these. Unfortunately the date on these particular files are formatted as "yyyyMMdd" and I need to covert the date into format "yyyy-MM-dd". I have been trying to use the offset function to get particular index positions, and I have found several examples of how you would return the offset of particular digits in the string, example:
set theposition to offset of 10 in theString -- this works
(which could return 5 or 7) but I have not found examples of how to call the digits at a specific index:
set _day to offset 7 of file_date_raw -- error
"Finder got an error: Some parameter is missing for offset." number -1701
How would you do this, or is there a totally better way I'm unaware of?
To "call the digits at a specific index", you use:
text 1 thru 4 of myString
If you know that each string has 8 characters in the yyyymmdd format, then you don't need to use 'offset' or any parsing, just add in the -'s, using text x thru y to dissect the string.
set d to "20011018"
set newString to (text 1 thru 4 of d) & "-" & (text 5 thru 6 of d) & "-" & (text 7 thru 8 of d)
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)]