This question already has answers here:
What is the reduce() function doing, in Swift
(4 answers)
Closed 9 months ago.
I am reading iOS 13 Programming Fundamentals with Swift, got to the part about reduce() and I think I understand it more or less, but then there is reduce(into:) and this piece of code:
let nums = [1,2,3,4,5]
let result = nums.reduce(into: [[],[]]) { temp, i in
temp[i%2].append(i)
}
// result is now [[2,4],[1,3,5]]
So this code takes an array of Int and splits it into 2 arrays, even and odd. The problem is that I have no idea what's happening inside the brackets {}.
In the case of reduce, the first parameter is the first one of the iteration and then the closure is supposed to process all the items one after the other, similar to map() but more powerful (here one loop is enough to get the two arrays but with map() I would need 2 loops, according to the book).
I cannot understand the syntax here anyway, especially what does "temp" stand for and that use of "in". And how is "append()" appending the value to the proper array??
Inside the closure, "temp" is the result format which is [[][]] and "i" is each number. As you said it processes all numbers in a loop. When % is used it returns the division remainder, so for the odd numbers like "1,3,5", it returns "1" and for the even numbers "0", which means that "temp" appends these values to the array in these respective indexes.
So if we debug and replace the variables for constants the results would be:
temp[1].append(1) //1%2 = 1/2 left 1 [[][1]]
temp[0].append(2) //2%2 = 2/2 left 0 [[2][1]]
temp[1].append(3) //3%2 = 3/2 = 1 left 1 [[2][1,3]]
temp[0].append(4) //4%2 = 4/2 left 0 [[2,4][1,3]]
temp[1].append(5) //5%2 = 5/2 = 2 left 1 [[2,4][1,3,5]]
According to the documentation the closure is called sequentially with a mutable accumulating value initialized that when exhausted, is returned to the caller.
Related
I recently began to study swift as my first programming language and there's a very simple term that i can't find an answer for it.
take a look at this code:
var a = [Int]()
repeat {
let randomNumbers = Int.random(in: 0...10)
if a.contains(randomNumbers) == false {
a.append(randomNumbers)
}
print(randomNumbers)
} while (a.count < 10)
so this code is supposed to add 10 numbers (no duplicates) from 0...10 into the array until all unique integers are listed. what I don't understand is the role of "while" here.
doesn't the last line mean the number of "generated numbers" must be less than 10? then why every time I run the code I get more than 10 numbers (say 30-40) in the console?
Also according to the code, this code must not generate dupes. then why do I get some numbers printed 2,3 times in the console?
you check if randomNumbers already added to array a, if yes, the loop will be executed again, until a.count < 10.
if you move the print statement inside the if statement, it will be printed exactly 10 times.
I'm a little baffled about the inner work of the sequence expression in F#.
Normally if we make a sequential file reader with seq with no intentional caching of data
seq {
let mutable current = file.Read()
while current <> -1 do
yield current
}
We will end up with some weird behavior if we try to do some re-iterate or backtracking, My Idea of this was, since Read() is a function calling some mutable value we can't expect the output to be correct if we re-iterate. But then this behaves nicely even on boundary reading?
let Read path =
seq {
use fp = System.IO.File.OpenRead path
let buf = [| for _ in 0 .. 1024 -> 0uy |]
let mutable pos = 1
let mutable current = 0
while pos <> 0 do
if current = 0 then
pos <- fp.Read(buf, 0, 1024)
if pos > 0 && current < pos then
yield buf.[current]
current <- (current + 1) % 1024
}
let content = Read "some path"
We clearly use the same buffer to enhance performance, but assuming that we read the 1025 byte, it will trigger an update to the buffer, if we then try to read any byte with position < 1025 after we still get the correct output. How can that be and what are the difference?
Your question is a bit unclear, so I'll try to guess.
When you create a seq { }, you're essentially creating a state machine which will run only as far as it needs to. When you request the very first element from it, it'll start at the top and run until your first yield instruction. Then, when you request another value, it'll run from that point until the next yield, and so on.
Keep in mind that a seq { } produces an IEnumerable<'T>, which is like a "plan of execution". Each time you start to iterate the sequence (for example by calling Seq.head), a call to GetEnumerator is made behind the scenes, which causes a new IEnumerator<'T> to be created. It is the IEnumerator which does the actual providing of values. You can think of it in more classical terms as having an array over which you can iterate (an iterable or enumerable) and many pointers over that array, each of which are at different points in the array (many iterators or enumerators).
In your first code, file is most likely external to the seq block. This means that the file you are reading from is baked into the plan of execution; no matter how many times you start to iterate the sequence, you'll always be reading from the same file. This is obviously going to cause unpredictable behaviour.
However, in your second code, the file is opened as part of the seq block's definition. This means that you'll get a new file handle each time you iterate the sequence or, essentially, a new file handle per enumerator. The reason this code works is that you can't reverse an enumerator or iterate over it multiple times, not with a single thread at least.
(Now, if you were to manually get an enumerator and advance it over multiple threads, you'd probably run into problems very quickly. But that is a different topic.)
This question already has answers here:
Lua for loop reduce i? Weird behavior [duplicate]
(3 answers)
Closed 7 years ago.
im trying this in lua:
for i = 1, 10,1 do
print(i)
i = i+2
end
I would expect the following output:
1,4,7,10
However, it seems like i is getting not affected, so it gives me:
1,2,3,4,5,6,7,8,9,10
Can someone tell my a bit about the background concept and what is the right way to modify the counter variable?
As Colonel Thirty Two said, there is no way to modify a loop variable in Lua. Or rather more to the point, the loop counter in Lua is hidden from you. The variable i in your case is merely a copy of the counter's current value. So changing it does nothing; it will be overwritten by the actual hidden counter every time the loop cycles.
When you write a for loop in Lua, it always means exactly what it says. This is good, since it makes it abundantly clear when you're doing looping over a fixed sequence (whether a count or a set of data) and when you're doing something more complicated.
for is for fixed loops; if you want dynamic looping, you must use a while loop. That way, the reader of the code is aware that looping is not fixed; that it's under your control.
When using a Numeric for loop, you can change the increment by the third value, in your example you set it to 1.
To see what I mean:
for i = 1,10,3 do
print(i)
end
However this isn't always a practical solution, because often times you'll only want to modify the loop variable under specific conditions. When you wish to do this, you can use a while loop (or if you want your code to run at least once, a repeat loop):
local i = 1
while i < 10 do
print(i)
i = i + 1
end
Using a while loop you have full control over the condition, and any variables (be they global or upvalues).
All answers / comments so far only suggested while loops; here's two more ways of working around this problem:
If you always have the same step size, which just isn't 1, you can explicitly give the step size as in for i =start,end,stepdo … end, e.g. for i = 1, 10, 3 do … or for i = 10, 1, -1 do …. If you need varying step sizes, that won't work.
A "problem" with while-loops is that you always have to manually increment your counter and forgetting this in a sub-branch easily leads to infinite loops. I've seen the following pattern a few times:
local diff = 0
for i = 1, n do
i = i+diff
if i > n then break end
-- code here
-- and to change i for the next round, do something like
if some_condition then
diff = diff + 1 -- skip 1 forward
end
end
This way, you cannot forget incrementing i, and you still have the adjusted i available in your code. The deltas are also kept in a separate variable, so scanning this for bugs is relatively easy. (i autoincrements so must work, any assignment to i below the loop body's first line is an error, check whether you are/n't assigning diff, check branches, …)
Suppose I have a factor variable with labels "a" "b" and "c" and want to see which observations have a label of "b". Stata refuses to parse
gen isb = myfactor == "b"
Sure, there is literally a "type mismatch", since my factor is encoded as an integer and so cannot be compared to the string "b". However, it wouldn't kill Stata to (i) perform the obvious parse or (ii) provide a translator function so I can write the comparison as label(myfactor) == "b". Using decode to (re)create a string variable defeats the purpose of encoding, which is to save space and make computations more efficient, right?
I hadn't really expected the comparison above to work, but I at least figured there would be a one- or two-line approach. Here is what I have found so far. There is a nice macro ("extended") function that maps the other way (from an integer to a label, seen below as local labi: label ...). Here's the solution using it:
// sample data
clear
input str5 mystr int mynum
a 5
b 5
b 6
c 4
end
encode mystr, gen(myfactor)
// first, how many groups are there?
by myfactor, sort: gen ng = _n == 1
replace ng = sum(ng)
scalar ng = ng[_N]
drop ng
// now, which code corresponds to "b"?
forvalues i = 1/`=ng'{
local labi: label myfactor `i'
if "b" == "`labi'" {
scalar bcode = `i'
break
}
}
di bcode
The second step is what irks me, but I'm sure there's a also faster, more idiomatic way of performing the first step. Can I grab the length of the label vector, for example?
An example:
clear all
set more off
sysuse auto
gen isdom = 1 if foreign == "Domestic":`:value label foreign'
list foreign isdom in 1/60
This creates a variable called isdom and it will equal 1 if foreigns's value label is equal to "Domestic". It uses an extended macro function.
From [U] 18.3.8 Macro expressions:
Also, typing
command that makes reference to `:extended macro function'
is equivalent to
local macroname : extended macro function
command that makes reference to `macroname'
This explains one of the two : in the offered syntax. The other can be explained by
... to specify value labels directly in an expression, rather than through
the underlying numeric value ... You specify the label in double quotes
(""), followed by a colon (:), followed by the name of the value
label.
The quote is from Stata tip 14: Using value labels in expressions, by Kenneth Higbee, The Stata Journal (2004). Freely available at http://www.stata-journal.com/sjpdf.html?articlenum=dm0009
Edit
On computing the number of distinct observations, another way is:
by myfactor, sort: gen ng = _n == 1
count if ng
scalar sc_ng = r(N)
display sc_ng
But yours is fine. In fact, it is documented here: http://www.stata.com/support/faqs/data-management/number-of-distinct-observations/, along with more methods and comments.
In section 4, Tables, in The Implementation of Lua 5.0 there is and example:
local t = {100, 200, 300, x = 9.3}
So we have t[4] == nil. If I write t[0] = 0, this will go to hash part.
If I write t[5] = 500 where it will go? Array part or hash part?
I would eager to hear answer for Lua 5.1, Lua 5.2 and LuaJIT 2 implementation if there is difference.
Contiguous integer keys starting from 1 always go in the array part.
Keys that are not positive integers always go in the hash part.
Other than that, it is unspecified, so you cannot predict where t[5] will be stored according to the spec (and it may or may not move between the two, for example if you create then delete t[4].)
LuaJIT 2 is slightly different - it will also store t[0] in the array part.
If you need it to be predictable (which is probably a design smell), stick to pure-array tables (contiguous integer keys starting from 1 - if you want to leave gap use a value of false instead of nil) or pure hash tables (avoid non-negative integer keys.)
Quoting from Implementation of Lua 5.0
The array part tries to store the values corresponding to integer keys from 1 to some limit n.Values corresponding to non-integer keys or to integer keys outside the array range are
stored in the hash part.
The index of the array part starts from 1, that's why t[0] = 0 will go to hash part.
The computed size of the array part is the largest nsuch that at least half the slots between 1 and n are in use (to avoid wasting space with sparse arrays) and there is at least one used slot between n/2+1 and n(to avoid a size n when n/2 would do).
According from this rule, in the example table:
local t = {100, 200, 300, x = 9.3}
The array part which holds 3 elements, may have a size of 3, 4 or 5. (EDIT: the size should be 4, see #dualed's comment.)
Assume that the array has a size of 4, when writing t[5] = 500, the array part can no longer hold the element t[5], what if the array part resize to 8? With a size of 8, the array part holds 4 elements, which is equal to (so, not less that) half of the array size. And the index from between n/2+1 and n, which in this case, is 5 to 8, has one element:t[5]. So an array size of 8 can accomplish the requirement. In this case, t[5] will go to the array part.