Getting data elements through match

row.match in r
match data frame columns r
r match rows in two data frames
index match in r
match columns in r
r match order
matching variables in r
match column values in r

I have an a list of data with the format [(StudID,Int)] where type StudID = String. Hence, I have a set of data:

marks = [("8",90),("10",100),("5",86),("3",45)]

I now want to obtain the second value based on a matching ID which is the first element in each tuple. For example, if the ID I'm looking for is "5" then I want the value of 86 and so on. I tried something as below:

getStudMarks :: StudID -> [(StudID,Int)] -> Int
getStudMarks studid [(id,mark):xs] = if studid == id then mark else getStudMarks studid xs

But I'm getting thrown

Couldn't match expected type ‘(StudID, Int)’ with actual type ‘[(StudID, Int)]’

highlighting the [(id,mark):xs] section in my code

Am I doing the recursion correctly?

(id, mark):xs is the list you want to pattern-match on; you are wrapping that in another layer of brackets that implies a value of type [[(StudID, Int)]].

getStudMarks studid ((id,mark):xs) = if studid == id then mark else getStudMarks studid marks.

However, you also need to consider what happens if getStudMarks ever receives an empty list as its second argument, which will happen if studid is never found.

getStudMarks _ [] = ???

Compare your function to lookup :: Eq => a -> [(a,b)] -> Maybe b (which, incidentally is all you really need here; change the type of getStudMarks, write getStudMarks = lookup, and you're done).

For this, we retrieve data elements contained in the answer graph matches, and join them along the query edges. Example 6. Fig. 6b depicts a query, which asks​  To enter the MATCH function and arguments: Open a blank Excel worksheet and enter the data in columns C, D, and E, as shown in the image below. Leave cell D2 blank, as that particular cell will host the function. Select cell D2 to make it the active cell.

Have you had a look at list comprehension?

getStudMarks studentId marks = [ id | (student, id) <- marks, student == studentId ]

Fact: Any model matches a set of data with no element. 2. Rule: A model matches a nonempty set of data if a. the first element of the model matches the first  Using MATCH The MATCH function is doesn’t return the value of data to you; you provide the value that you’re looking for, and the function returns the position of that value. It’s like asking where is #135 Main Street, and getting the answer that it’s the 4 th building down the street.

The recursion should be

getStudMarks studid [] = error "No value for id"
getStudMarks studid ((id,mark):xs) 
    | studid == id = mark
    | otherwise = getStudMarks studid xs

Other possibility

getStudMarks id [] = error "No value for id"
getStudMarks studid ((id,mark):xs) = if studid == id then mark else getStudMarks studId xs

Inspired by LearnYouAHaskell

If there is a match with one of the data-element names (line 212), control is transferred to statement 80 to begin processing the change in data value for that data  Getting Data from another worksheet (using INDEX & MATCH) You’re seeing a worksheet (‘Sheet1’) below with some dummy data. In the following image, you are seeing my ‘Get’ worksheet data. In the B2 cell, you see I have input value 3/2018.

In this case, the biological assay is gene expression and data was generated in documentation this operator is well-used and convenient once you get the hang input to the left and will evaluate each element to see if there is a match in the  Using an approximate match, searches for the value 1 in column A, finds the largest value less than or equal to 1 in column A, which is 0.946, and then returns the value from column C in the same row. 100 =VLOOKUP(0.7,A2:C10,3,FALSE) Using an exact match, searches for the value 0.7 in column A.

Probabilistic matching was performed using Integrity (Ascential Software, Data elements were available in all states except as noted: in Washington, only names was considered a match, and any record pair receiving a weight equal to or. The .data () method allows us to attach data of any type to DOM elements in a way that is safe from circular references and therefore from memory leaks. We can set several distinct values for a single element and retrieve them later: Using the data () method to update data does not affect attributes in the DOM.

What is a sustainable strategy for data matching? How do we know that the algorithms we put in place are still The business question is whether there are enough data elements that match that you'd reasonably consider  The MatchCollection object is populated as needed on a match-by-match basis. It is equivalent to the regular expression engine calling the Regex.Match method repeatedly and adding each match to the collection. This technique is used when the collection is accessed through its GetEnumerator method, or when it is accessed using the foreach

  • See lookup :: Eq a => a -> [(a,b)] -> Maybe b. Currently, you aren't dealing with the possibility that studid isn't found in the list.
  • Why don't use Data.Map? It is efficient than recursive function. To find a element in Map, just O(log n) instead of O(n)
  • I'm only wrapping it to one layer of [] bracket [(StudID,Int)] instead of [[(StudID,Int)]]. And what do you mean by getStudMarks = lookup. I didn't understood that part
  • (id, mark):xs is already a list, defined by the (:) constructor. Wrapping it in [...] creates a list whose sole element is the matched list; it's identical to ((id,mark):xs):[].
  • lookup (which is already defined in the Prelude) is basically the function you are trying to define.
  • there seems to be an error going on in studId((id,mark):xs) = getStudMarks studId xs. Shouldnt it be id instead? I'm confused
  • I'm more familiar with the guards syntax in your first example. Would it be possible to cater for empty list [] in the guards statement?
  • You can also use the multiple notation in combination with the guards, so first line of second option combined with first option. On the other hand the second option is how usually Haskell is written.
  • It should not be getStudMarks id ((id,mark):xs) instead of getStudMarks studId ((id,mark):xs) in the last line, because this case is covered in the second line of the second option.
  • getStudMarks id ((id,mark):xs) = mark is not valid Haskell.