Related
Let's say I have a vector, the values are from 1 to 10. I want that if you find 5 and 5 next to each other, remove them together with the next elements.
input
[1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10]
expected output
[1, 2, 3, 4]
This was my attempt. I'm finding index to remove, but borrowing rules are making me stuck.
let mut element = vec![1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10];
for (index, val) in element.iter().enumerate() {
if *val == 5 {
if let Some(next_val) = element.get(index + 1) {
if *next_val == 5 {
//element.drain(index..);
}
}
}
}
Rust is saving you from iterator invalidation (a common source of bugs in other languages). This is an error that usually happens when you try to modify a data structure while concurrently iterating over it. You cannot move on to the (now-deleted) next element after calling element.drain(index..). So you need to add a break after that point to avoid memory unsafety.
In this case just adding break; is sufficient to make the code compile. However, for a more concise, linear solution, take full advantage of the iterators and methods provided by the standard library:
if let Some(index) = element.windows(2).position(|pair| pair[0] == pair[1]) {
element.truncate(index);
}
windows(2) on a slice gives an iterator over subslices of length 2, and the position call returns the index of the first element of that iterator for which the two elements of the slice are equal. (If no such pair exists, position returns None.)
I find that the position closure becomes more obvious with the (currently unstable) array_windows feature:
if let Some(index) = element.array_windows().position(|[x, y]| x == y) {
element.truncate(index);
}
Playground
Related
is it possible to filter on a vector in-place?
You can't do what you want to do because you want to remove some elements from a vector while you are iterating it. And this is a big mistake. Note that removing any elements from a vector invalidates the iterators, hence you will access unexpected locations so rust doesn't allow UBs
You can use something like the following
let mut elements = vec![1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10];
let mut first_repeated_five_index_op = None;
for index in 0..elements.len()-1{
if elements[index] == 5 && elements[index + 1] == 5{
first_repeated_five_index_op = Some(index);
break;
}
}
if let Some(first_repeated_five_index) = first_repeated_five_index_op{
elements.truncate(first_repeated_five_index);
}
println!("{:?}", elements);
See a Demo
I have a random array of numbers 1 to five occurring in ramdom sometimes [1,1,1,1,2,2] etc. I am tasked with finding the value with highest occurrence all the the time regardless. I achieved that in javascript like below using a library called ramda here . After reading the documentation, i went with a solution like below.
// filter out duplication in array that way you can get the uniq represented numbers
const uniqueItems = R.uniq(params);
// use the unique numbers as keys and create a new array of object
const mappedItemsWithRepresentations = map((a) => ({ color: a, rep: params.filter(b => b === a).length }), uniqueItems);
// and then finally, select the item with highest rep and return it key
const maxRepItem = mappedItemsWithRepresentations.reduce((acc, curr) => acc.rep > curr.rep ? acc : curr, []);
return maxRepItem.key; // gives me the correct value i need
However, reading through more in the documentation and going through the example here, i realised there is a way i can combine the logic above and simply with ramda. I tried numerous attempt possible and the closest i could get are below.
const getMaxRep = curry(pipe(uniq, map((a) => ({ color: a, rep: filter(b => b === a).length })), pipe(max(pathEq("rep")), tap(console.log))));
console.log("Max Rep here", getMaxRep(params));
I also tried utilising the reduced feature here, all to no avail. Please how do i arrange achieve that ? Any help will be appreciated.
Ramda has R.countBy to get the number of occurrences. You can convert the resulting object of country to pairs [value, count], and then reduce it to find the pair with the highest count:
const { pipe, countBy, identity, toPairs, reduce, maxBy, last, head } = R
const fn = pipe(
countBy(identity), // count the occurrences
toPairs, // convert to pairs of [value, count]
reduce(maxBy(last), [0, 0]), // reduce to find the maximum occurrence
head, // get the actual value
Number, // convert back to an number
)
const arr = [1,1,1,1,2,2]
const result = fn(arr)
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script>
A slight variation on this idea that collects values with the same count to an array. This will handle cases in which the frequency of several items is identical:
const { pipe, countBy, identity, toPairs, invert, reduce, maxBy, last, head, map } = R
const fn = pipe(
countBy(identity), // count the occurrences
invert, // combine all values with the same count
toPairs, // convert to pairs of [value, count]
reduce(maxBy(head), [0, 0]), // reduce to find the maximum occurrence
last, // get the actual values
map(Number), // convert back to numbers
)
const arr = [1,1,1,1,2,2,3,3,3,3]
const result = fn(arr)
console.log(result)
<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script>
nice use case, try this:
const maxReduce = reduce(maxBy(last), [0,0])
const getMaxRep = pipe(countBy(identity), toPairs, maxReduce, head)
console.log(getMaxRep([1,1,1,1,2,2]))
countBy is a really nice start, sadly Ramda don't support reduce for object but we can convert to an array of arrays using toPairs function and finish the work.
It's not entirely clear to me what it is you're asking for.
But it might be something like this:
const maxRep = pipe (
countBy (identity),
toPairs,
map (zipObj(['color', 'rep'])),
reduce (maxBy (prop ('rep')), {rep: -Infinity}),
)
const params = [1, 2, 3, 4, 2, 3, 5, 2, 3, 2, 1, 1, 4, 5, 5, 3, 2, 5, 1, 5, 2]
console .log (
maxRep (params)
)
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.27.0/ramda.js"></script>
<script> const {pipe, countBy, identity, toPairs, map, zipObj, reduce, maxBy, prop} = R </script>
We start with a list of values drawn from {1, 2, 3, 4, 5}, occuring in some random, multiply-occuring order.
With countBy(identity) we change the original list into something like
{"1": 4, "2": 6, "3": 4, "4": 2, "5": 5}
with the counts associated with each entry.
toPairs formats that as an array like
[["1", 4], ["2", 6], ["3", 4], ["4", 2], ["5", 5]]
(You could also use Object.entries here.)
Then by calling map (zipObj (['color', 'rep'])), we turn this into
[{"color": "1", "rep": 4}, {"color": "2", "rep": 6}, ...]
Finally, we reduce the result, using maxBy (prop ('rep')), which chooses the one with the maximum rep value. For the initial value to the max call, we create a dummy object, {rep: -Infinity} that will compare less than any in your list.
If you wanted to also keep that final intermediate structure, you could rename that function to makeReps, dropping off the last function in the pipeline, and then making a new maxRep out of it.
Then you could call
const reps = makeResps (params)
const maxVal = maxRep (reps)
and use both.
But all this presupposes that the value with color and rep properties is what you need. If you just need the count then the other solutions already here handle that fine.
I have to give list of values into in clause of SQL query but while retrieving the values [ ] also come along with data which is not readable by query language.
For example I have list as:
def val = new ArrayList<Integer>(Arrays.asList(1,2,3,4,5,6,7,8))
while doing println(val) output come as [1, 2, 3, 4, 5, 6, 7, 8] but in the query it is needed as: 1, 2, 3, 4, 5, 6, 7, 8
In java this one works as System.out.println(val.toString().replaceAll("[\\[\\]]", "")) but not in groovy. Can't we use collection to remove like this?
Instead of:
def val = new ArrayList(Arrays.asList(1,2,3,4,5,6,7,8))
use:
def val = new ArrayList(Arrays.asList(1,2,3,4,5,6,7,8)).join(', ')
or simply:
def val = [1,2,3,4,5,6,7,8].join(', ')
Try using g-strings and the minus operator:
println "${val}" - '[' - ']'
I am a little confused on the answer that Xcode is giving me to this experiment in the Swift Programming Language Guide:
// Use a for-in to iterate through a dictionary (experiment)
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25]
]
var largest = 0
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
largest = number
}
}
}
largest
I understand that as the dictionary is being transversed, the largest number is being set to the variable, largest. However, I am confused as to why Xcode is saying that largest is being set 5 times, or 1 time, or 3 times, depending on each test.
When looking through the code, I see that it should be set 6 times in "Prime" alone (2, 3, 5, 7, 11, 13). Then it should skip over any numbers in "Fibonacci" since those are all less than the largest, which is currently set to 13 from "Prime". Then, it should be set to 16, and finally 25 in "Square", yielding a total of 8 times.
Am I missing something entirely obvious?
Dictionaries in Swift (and other languages) are not ordered. When you iterate through the dictionary, there's no guarantee that the order will match the initialization order. In this example, Swift processes the "Square" key before the others. You can see this by adding a print statement to the loop. 25 is the 5th element of Square so largest would be set 5 times for the 5 elements in Square and then would stay at 25.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25]
]
var largest = 0
for (kind, numbers) in interestingNumbers {
println("kind: \(kind)")
for number in numbers {
if number > largest {
largest = number
}
}
}
largest
This prints:
kind: Square
kind: Prime
kind: Fibonacci
let dict : [String : Any] = ["FirstName" : "Maninder" , "LastName" : "Singh" , "Address" : "Chandigarh"]
dict.forEach { print($0) }
Result would be
("FirstName", "Maninder")
("LastName", "Singh")
("Address", "Chandigarh")
This is a user-defined function to iterate through a dictionary:
func findDic(dict: [String: String]) {
for (key, value) in dict {
print("\(key) : \(value)")
}
}
findDic(dict: ["Animal": "Lion", "Bird": "Sparrow"])
// prints…
// Animal : Lion
// Bird : Sparrow
If you want to iterate over all the values:
dict.values.forEach { value in
// print(value)
}
Here is an alternative for that experiment (Swift 3.0). This tells you exactly which kind of number was the largest.
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var whichKind: String? = nil
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
whichKind = kind
largest = number
}
}
}
print(whichKind)
print(largest)
OUTPUT:
Optional("Square")
25
You can also use values.makeIterator() to iterate over dict values, like this:
for sb in sbItems.values.makeIterator(){
// do something with your sb item..
print(sb)
}
You can also do the iteration like this, in a more swifty style:
sbItems.values.makeIterator().forEach{
// $0 is your dict value..
print($0)
}
sbItems is dict of type [String : NSManagedObject]
I have a set of datapoints such as (THIS IS AN EXAMPLE)
val=4; (*this value is calculated before in the program, so it not known a priori*)
x={0,1,2,3};
data=Table[0, {val}];
data[[1]] = {1,5,6,8};
data[[2]] = {9,7,1,3};
data[[3]] = {3,4,5,6};
data[[4]] = {2,2,4,6};
Now I can plot each of these data with ListPlot as
ListPlot[Transpose[{x,data[[1]]}]]
and if I want to plot more than one I can do
ListPlot[{Transpose[{x, data[[1]]}], Transpose[{x, data[[2]]}]}]
but how can I plot all of them in one code single line, by considering that val is calculated before in the program?
Is there a way to do something like
For[i = 1, i < val + 1, i++, ListPlot[Transpose[{x,data[i]}]]......]
having a single graph with all x-y curves?
Indeed I would like a static picture of
Manipulate[ListPlot[Transpose[{x, data[[i]]}]], {i, 1, val,1}]
Thanks
Virgilio
You want to "do the same thing" to every element of a list. That should tell you to think of using Map. Your list is named data and each element is your four element sublist. If you look at the help page for Map it shows you need to think up a function that does what you need to do to each individual sublist. You have already understood that you need to use Transpose with x and your sublist so that tells you your function and you are almost there. The result of Map will be a list of all those results. So
In[1]:= x = {0, 1, 2, 3};
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListPlot[Map[Transpose[{x, #}] &, data], Joined -> True]
Out[3]= ...FourOverlaidPlotsSnipped...
Go through that a word at a time until you can really understand the thinking that was done to be able to write that. You will use this idea again and again if you keep using Mathematica.
For the example you give the cleanest method is to use DataRange:
data = {{1, 5, 6, 8}, {9, 7, 1, 3}, {3, 4, 5, 6}, {2, 2, 4, 6}};
ListLinePlot[data, DataRange -> {0, 3}]
Please ask your future questions on the dedicated Mathematica StackExchange site: