I'm trying to understand why behavior for the match regex is different from the behavior for find, from documentation here.
I have the following for match:
use regex::Regex;
{
let meow = String::from("This is a long string that I am testing regexes on in rust.");
let re = Regex::new("I").unwrap();
let x = re.is_match(&meow);
dbg!(x)
}
And get:
[src/lib.rs:142] x = true
Great, now let's identify the location of the match:
{
let meow = String::from("This is a long string that I am testing regexes on in rust.");
let re = Regex::new("I").unwrap();
let x = re.find(&meow).unwrap();
dbg!(x)
}
And I get:
let x = re.find(&meow).unwrap();
^^^^^ borrowed value does not live long enough
}
^ `meow` dropped here while still borrowed
`meow` does not live long enough
I think I'm following the documentation. Why does the string meow live long enough for a match but not long enough for find?
Writing a value without ; at the end of a { } scope effectively returns that value out of the scope. For example:
fn main() {
let x = {
let y = 10;
y + 1
};
dbg!(x);
}
[src/main.rs:7] x = 11
Here, because we don't write a ; after the y + 1, it gets returned from the inner scope and written to x.
If you write a ; after it, you will get something different:
fn main() {
let x = {
let y = 10;
y + 1;
};
dbg!(x);
}
[src/main.rs:7] x = ()
Here you can see that the ; now prevents the value from being returned. Because no value gets returned from the inner scope, it implicitly gets the empty return type (), which gets stored in x.
The same happens in your code:
use regex::Regex;
fn main() {
let z = {
let meow = String::from("This is a long string that I am testing regexes on in rust.");
let re = Regex::new("I").unwrap();
let x = re.is_match(&meow);
dbg!(x)
};
dbg!(z);
}
[src/main.rs:9] x = true
[src/main.rs:12] z = true
Because you don't write a ; after the dbg!() statement, its return value gets returned from the inner scope. The dbg!() statement simply returns the value that gets passed to it, so the return value of the inner scope is x. And because x is just a bool, it gets returned without a problem.
Now let's look at your second example:
use regex::Regex;
fn main() {
let z = {
let meow = String::from("This is a long string that I am testing regexes on in rust.");
let re = Regex::new("I").unwrap();
let x = re.find(&meow).unwrap();
dbg!(x)
};
dbg!(z);
}
error[E0597]: `meow` does not live long enough
--> src/main.rs:8:25
|
4 | let z = {
| - borrow later stored here
...
8 | let x = re.find(&meow).unwrap();
| ^^^^^ borrowed value does not live long enough
9 | dbg!(x)
10 | };
| - `meow` dropped here while still borrowed
And now it should be more obvious what's happening: It's basically the same as the previous example, just that the returned x is now a type that internally borrows meow. And because meow gets destroyed at the end of the scope, x cannot be returned, as it would outlive meow.
The reason why x borrows from meow is because regular expression Matches don't actually copy the data they matched, they just store a reference to it.
So if you add a ;, you prevent the value from being returned from the scope, changing the scope return value to ():
use regex::Regex;
fn main() {
let z = {
let meow = String::from("This is a long string that I am testing regexes on in rust.");
let re = Regex::new("I").unwrap();
let x = re.find(&meow).unwrap();
dbg!(x);
};
dbg!(z);
}
[src/main.rs:9] x = Match {
text: "This is a long string that I am testing regexes on in rust.",
start: 27,
end: 28,
}
[src/main.rs:12] z = ()
Brand new to haskell and I need to print out the data contained on a seperate row for each individual item
Unsure on how to
type ItemDescr = String
type ItemYear = Int
type ItemPrice = Int
type ItemSold = Int
type ItemSales = Int
type Item = (ItemRegion,ItemDescr,ItemYear,ItemPrice,ItemSold,ItemSales)
type ListItems = [Item]
rownumber x
| x == 1 = ("Scotland","Desktop",2017,900,25,22500)
| x == 2 = ("England","Laptop",2017,1100,75,82500)
| x == 3 = ("Wales","Printer",2017,120,15,1800)
| x == 4 = ("England","Printer",2017,120,60,7200)
| x == 5 = ("England","Desktop",2017,900,50,45000)
| x == 6 = ("Wales","Desktop",2017,900,20,18000)
| x == 7 = ("Scotland","Printer",2017,25,25,3000)
showall
--print??
So for example on each individual line
show
"Scotland","Desktop",2017,900,25,22500
followed by the next record
Tip 1:
Store the data like this
items = [("Scotland","Desktop",2017,900,25,22500),
("England","Laptop",2017,1100,75,82500),
("Wales","Printer",2017,120,15,1800),
("England","Printer",2017,120,60,7200),
("England","Desktop",2017,900,50,45000),
("Wales","Desktop",2017,900,20,18000),
("Scotland","Printer",2017,25,25,3000)]
Tip 2:
Implement this function
toString :: Item -> String
toString = undefined -- do this yourselves
Tip 3:
Try to combine the following functions
unlines, already in the Prelude
toString, you just wrote it
map, does not need any explanation
putStrLn, not even sure if this is a real function, but you need it anyway.
($), you can do without this one, but it will give you bonus points
I have a function that uses two colomns in a dataframe:
def create_time(var, var1):
if var == "Helår":
y = var1+'Q4'
else:
if var == 'Halvår':
y = var1+'Q2'
else:
y = var1+'Q'+str(var)[0:1]
return y
Now i want to loop hrough my dataframe, creatring a new column using the function, where var and var1 are columns in the dataframe
I try with the following, but have no luck:
for row in bd.iterrows():
A = str(bd['Var'])
B = str(bd['Var1'])
bd['period']=create_time(A,B)
Looping is a last resort. There is usually a "vectorized" way to operate on the entire DataFrame, which always faster and usually more readable too.
To apply your custom function to each row, use apply with the keyword argument axis=1.
bd['period'] = bd[['Var', 'Var1']].apply(lambda x: create_time(*x), axis=1)
You might wonder why it's not just bd.apply(create_time). Since create_time wants two arguments, we have to "unpack" the row x into its two values and pass those to the function.
val x =10;
fun power (x:int, y:int) =
if y=1
then x
else
x * power (x,y-1)
val z = power 2,3
It gives me an error Replacing COMMA with EQUALOP . I dont understand whats the error in the code ??
You need parentheses around the argument to power in the declaration of z:
val z = power (2,3);
I have this SML code. I don't know why I cannot compile this :
fun score =
let
val sum = 3; (* error at this line : SYNTAX ERROR : inserting LPAREN *)
if sum div 2 > 0
then sum = 0
else sum = 1
(*some other code*)
in
sum (* I want to return sum after some steps of calculation *)
end
There are more issues with your code, than jacobm points out.
You are also missing a function argument. Functions in SML always takes one argument. For example
fun score () =
let val sum = 3
val sum = if sum div 2 > 0
then sum = 0
else sum = 1
in
sum
end
However this still doesn't make much sense. since the expressions sum = 0 and sum = 1 evaluates to a Boolean.
A let-expression is used to make some local declarations which are only visible inside the in ... end part. Thus the calculations you wan't to do with sum, should probably be done inside the in ... end part, unless you wan't to express it as a means of a function.
One such example is
fun score () =
let val sum = 3
in
if sum div 2 > 0
then ...
else ...
end
If we look at the syntax of a let-expression, it probably makes more sense
let
<declaration>
in
<expr> ; ... ; <expr>
end
Since if-then-else is an expression, it can't be in the "declarations part" by itself.
That syntax just isn't legal -- in between let and in all you're allowed to have is a series of val name = expr fragments. You can do this, though:
fun score =
let val sum = 3
val sum = if sum div 2 > 0
then sum = 0
else sum = 1
in
sum
end
I would consider it a bit of a weird style to use sum for both variable names, but it's legal.