In Crystal, are there any alternatives to the following, but without creating the value of b after the end?
my_val = begin
a = 1
b = 2
a + b
end
b should be undefined here.
I release it's possible to use -> {}.call (but this creates an unneeded closure). There's also 1.times { ... }, but this seems hacky. Is there a shortcut to this within the language itself that is idiomatic?
For your particular case the following could be not idiomatic, but depending on what you want to achieve this can be good to be aware of:
Macro
myval = {% begin %}
a = 1
b = 2
a + b
{% end %}
myval #=> 3
b #=> undefined local variable or method 'b' (did you mean 'p'?)
Method
def myval
a = 1
b = 2
a + b
end
myval #=> 3
b #=> undefined local variable or method 'b' (did you mean 'p'?)
Block
myval = :any.try do
a = 1
b = 2
a + b
end
myval #=> 3
b #=> undefined local variable or method 'b' (did you mean 'p'?)
Related
I need to create a function that replaces a letter with the letter 13 letters after it in the alphabet (without using encode). I'm relatively new to Python so it has taken me a while to figure out a way to do this without using Encode.
Here's what I have so far. When I use this to type in a normal word like "hello" it works but if I pass through a sentence with special characters I can't figure out how to JUST include letters of the alphabet and skip numbers, spaces or special characters completely.
def rot13(b):
b = b.lower()
a = [chr(i) for i in range(ord('a'),ord('z')+1)]
c = []
d = []
x = a[0:13]
for i in b:
c.append(a.index(i))
for i in c:
if i <= 13:
d.append(a[i::13][1])
elif i > 13:
y = len(a[i:])
z = len(x)- y
d.append(a[z::13][0])
e = ''.join(d)
return e
EDIT
I tried using .isalpha() but this doesn't seem to be working for me - characters are duplicating for some reason when I use it. Is the following format correct:
def rot13(b):
b1 = b.lower()
a = [chr(i) for i in range(ord('a'),ord('z')+1)]
c = []
d = []
x = a[0:13]
for i in b1:
if i.isalpha():
c.append(a.index(i))
for i in c:
if i <= 12:
d.append(a[i::13][1])
elif i > 12:
y = len(a[i:])
z = len(x)- y
d.append(a[z::13][0])
else:
d.append(i)
if message[0].istitle() == True:
d[0] = d[0].upper()
e = ''.join(d)
return e
Following on from comments. OP was advised to use isalpha, and wondering why that's causing duplication (see OP's edit)
This isn't tied to the use of isalpha, it's to do with the second for loop
for i in c:
isn't necessary, and is causing the duplication. You should remove that. Instead you can do the same by just using index = a.index(i). You were already doing this, but for some reason appending to a list instead and causing confusion
Use the index variable any time you would have used i inside the for i in c loop. On a side note, in nested for loops try not to reuse the same variables. It just causes confusion...but that's a matter for code review
Assuming you do all that right it should work.
Is there any such thing as static variables in Ruby that would behave like they do in C functions?
Here's a quick example of what I mean. It prints "6\n7\n" to the console.
#include <stdio.h>
int test() {
static int a = 5;
a++;
return a;
}
int main() {
printf("%d\n", test());
printf("%d\n", test());
return 0;
}
What I understand about this C code is that the static variable is initialized with a value (5 in this case), and its value is persisted across function calls.
In Ruby and languages with the same level of abstraction, this same thing is typically achieved either by using a variable that sits outside the scope of the function or by using an object to hold that variable.
def test()
#a ||= 5 # If not set, set it. We need to use an instance variable to persist the variable across calls.
#a += 1 # sum 1 and return it's value
end
def main()
puts test
puts test
0
end
Ruby's quirk is that you can use instance variables even outside of a class definition.
Similar to nicooga's answer but more self-contained:
def some_method
#var ||= 0
#var += 1
puts #var
end
Scope your variable in a method and return lambda
def counter
count = 0
lambda{count = count+1}
end
test = counter
test[]
#=>1
test[]
#=>2
You can use a global variable:
$a = 5
def test
$a += 1
end
p test #=> 6
p test #=> 7
Use a variable in the singleton class (The static class itself)
class Single
def self.counter
if #count
#count+=1
else
#count = 5
end
end
end
In ruby, any class is an object which has only one instance.
Therefore, you can create an instance variable on the class, and it will work as a "static" method ;)
Output:
ruby 2.5.5p157 (2019-03-15 revision 67260) [x86_64-linux]
=> :counter
Single.counter
=> 5
Single.counter
=> 6
Single.counter
=> 7
To get this behaviour on the main scope, you can do:
module Countable
def counter
if #count
#count+=1
else
#count = 5
end
end
end
=> :counter
self.class.prepend Countable # this "adds" the method to the main object
=> Object
counter
=> 5
counter
=> 6
counter
=> 7
I think a standard way to do this is to use Fiber.
f = Fiber.new do
a = 5
loop{Fiber.yield a += 1}
end
puts f.resume # => 6
puts f.resume # => 7
...
To populate missing data with a fixed range of values
I would like to check how to populate column aktype with a range of values (the range of values for the same pidlink are always fixed at 11 types of values listed below) for those cells with missing values. I have about 17,000+ observations that are missing.
The range of values are as follows:
A
B
C
D
E
G
H
I
J
K
L
I have tried the following command but it does not work:-
foreach x of varlist aktype=1/11 {
replace aktype = "A" in 1 if aktype==""
replace aktype = "B" in 2 if aktype==""
replace aktype = "C" in 3 if aktype==""
replace aktype = "D" in 4 if aktype==""
replace aktype = "E" in 5 if aktype==""
replace aktype = "G" in 6 if aktype==""
replace aktype = "H" in 7 if aktype==""
replace aktype = "I" in 8 if aktype==""
replace aktype = "J" in 9 if aktype==""
replace aktype = "K" in 10 if aktype==""
replace aktype = "L" in 11 if aktype==""
}
Would appreciate it if you could advise on the right command to use. Many thanks!
I would generate a variable AK that has letters A-K in positions 1-11 (and 12-22, and 23-33, and so on). The replace missing values with the value of this variable AK.
* generate data
clear
set obs 20
generate aktype = ""
replace aktype = "foo" in 1/1
replace aktype = "bar" in 10/12
* generate variable with letters A-K
generate AK = char(65 + mod(_n - 1, 11))
* fill missing values
replace aktype = AK if missing(aktype)
list
This yields the following.
. list
+-------------+
| aktype AK |
|-------------|
1. | foo A |
2. | B B |
3. | C C |
4. | D D |
5. | E E |
|-------------|
This first addresses the comment "it does not work".
Generally, in this kind of forum you should always be specific and say exactly what happens, namely where the code breaks down and what the result is (e.g. what error message you get). If necessary, add why that is not what is wanted.
Specifically, in this case Stata would get no further than
foreach x of varlist aktype=1/11
which is illegal (as well as unclear to Stata programmers).
You can loop over a varlist. In this case looping over a single variable aktype is legal. (It is usually pointless, but that's style, not syntax.) So this is legal:
foreach x of varlist aktype
By the way, you define x as the loop argument, but never refer to it inside the loop. That isn't illegal, but it is unusual.
You can also loop over a numlist, e.g.
foreach x of numlist 1/11
although
forval x = 1/11
is a more direct way of doing that. All this follows from the syntax diagrams for the commands concerned, where whatever is not explicitly allowed is forbidden.
On occasions when you need to loop over a varlist and a numlist you will need to use different syntax, but what is best depends on the precise problem.
Now second to the question: I can't see any kind of rule in the question for which values get assigned A through L, so can't advise positively.
I've wrote the following spec
"An IP4 address" should "belong to just one class" in {
val addrs = for {
a <- Gen.choose(0, 255)
b <- Gen.choose(0, 255)
c <- Gen.choose(0, 255)
d <- Gen.choose(0, 255)
} yield s"$a.$b.$c.$d"
forAll (addrs) { ip4s =>
var c: Int = 0
if (IP4_ClassA.unapply(ip4s).isDefined) c = c + 1
if (IP4_ClassB.unapply(ip4s).isDefined) c = c + 1
if (IP4_ClassC.unapply(ip4s).isDefined) c = c + 1
if (IP4_ClassD.unapply(ip4s).isDefined) c = c + 1
if (IP4_ClassE.unapply(ip4s).isDefined) c = c + 1
c should be (1)
}
}
That is very clear in its scope.
The test passes successfully but when I force it to fail (for example by commenting out one of the if statements) then ScalaCheck correctly reports the error but the message doesn't mention correctly the actual value used to evaluate the proposition. More specifically I get:
[info] An IP4 address
[info] - should belong to just one class *** FAILED ***
[info] TestFailedException was thrown during property evaluation.
[info] Message: 0 was not equal to 1
[info] Location: (NetSpec.scala:105)
[info] Occurred when passed generated values (
[info] arg0 = "" // 4 shrinks
[info] )
where you can see arg0 = "" // 4 shrinks doesn't show the value.
I've tried to add even a simple println statement to review the cases but the output appears to be trimmed. I get something like this
192.168.0.1
189.168.
189.
1
SOLUTION
import org.scalacheck.Prop.forAllNoShrink
import org.scalatest.prop.Checkers.check
"An IP4 address" should "belong to just one class" in {
val addrs = for {
a <- Gen.choose(0, 255)
b <- Gen.choose(0, 255)
c <- Gen.choose(0, 255)
d <- Gen.choose(0, 255)
} yield s"$a.$b.$c.$d"
check {
forAllNoShrink(addrs) { ip4s =>
var c: Int = 0
if (IP4.ClassA.unapply(ip4s).isDefined) c = c + 1
if (IP4.ClassB.unapply(ip4s).isDefined) c = c + 1
if (IP4.ClassC.unapply(ip4s).isDefined) c = c + 1
if (IP4.ClassD.unapply(ip4s).isDefined) c = c + 1
if (IP4.ClassE.unapply(ip4s).isDefined) c = c + 1
c == (1)
}
}
}
This is caused by ScalaCheck's test case simplification feature. ScalaCheck just sees that your generator produces a string value. Whenever it finds a value that makes your property false, it tries to simplify that value. In your case, it simplifies it four times until it ends up with an empty string, that still makes your property false.
So this is expected, although confusing, behavior. But you can improve the situation in three different ways.
You can select another data structure to represent your IP addresses. This will make ScalaCheck able to simplify your test cases in a more intelligent way. For example, use the following generator:
val addrs = Gen.listOfN(4, Gen.choose(0,255))
Now ScalaCheck knows that your generator only produces lists of length 4, and that it only contains numbers between 0 and 255. The test case simplification process will take this into account and not create any values that couldn't have been produced by the generator from start. You can do the conversion to string inside your property instead.
A second method is to add a filter directly to your generator, which tells ScalaCheck how an IP address string should look like. This filter is used during test case simplification. Define a function that checks for valid strings and attach it to your existing generator this way:
def validIP(ip: String): Boolean = ...
val validAddrs = addrs.suchThat(validIP)
forAll(validAddrs) { ... }
The third method is to simply disable the test case simplification feature altogether by using forAllNoShrink instead of forAll:
Prop.forAllNoShrink(addrs) { ... }
I should also mention that the two first methods require ScalaCheck version >= 1.11.0 to function properly.
UPDATE:
The listOfN list length is actually not respected by the shrinker any more, due to https://github.com/rickynils/scalacheck/issues/89. Hopefully this can be fixed in a future version of ScalaCheck.
This is not homework, but an old exam question. I am curious to see the answer.
We are given an alphabet S={0,1,2,3,4,5,6,7,8,9,+}. Define the language L as the set of strings w from this alphabet such that w is in L if:
a) w is a number such as 42 or w is the (finite) sum of numbers such as 34 + 16 or 34 + 2 + 10
and
b) The number represented by w is divisible by 3.
Write a regular expression (and a DFA) for L.
This should work:
^(?:0|(?:(?:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\
+)*[369]0*)*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:
\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[
258](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0
\+)*[147])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+)(?:\+(?:0|(?:(?
:[369]|[147](?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)
*\+?(?:0\+)*[258])*(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]|0*(?:\+?(?:0\+)*
[369]0*)*\+?(?:0\+)*[147]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])|[258](?:0*(?
:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147])*
(?:0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[147]|0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)
*[258]0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*[258]))0*)+))*$
It works by having three states representing the sum of the digits so far modulo 3. It disallows leading zeros on numbers, and plus signs at the start and end of the string, as well as two consecutive plus signs.
Generation of regular expression and test bed:
a = r'0*(?:\+?(?:0\+)*[369]0*)*\+?(?:0\+)*'
b = r'a[147]'
c = r'a[258]'
r1 = '[369]|[147](?:bc)*(?:c|bb)|[258](?:cb)*(?:b|cc)'
r2 = '(?:0|(?:(?:' + r1 + ')0*)+)'
r3 = '^' + r2 + r'(?:\+' + r2 + ')*$'
r = r3.replace('b', b).replace('c', c).replace('a', a)
print r
# Test on 10000 examples.
import random, re
random.seed(1)
r = re.compile(r)
for _ in range(10000):
x = ''.join(random.choice('0123456789+') for j in range(random.randint(1,50)))
if re.search(r'(?:\+|^)(?:\+|0[0-9])|\+$', x):
valid = False
else:
valid = eval(x) % 3 == 0
result = re.match(r, x) is not None
if result != valid:
print 'Failed for ' + x
Note that my memory of DFA syntax is woefully out of date, so my answer is undoubtedly a little broken. Hopefully this gives you a general idea. I've chosen to ignore + completely. As AmirW states, abc+def and abcdef are the same for divisibility purposes.
Accept state is C.
A=1,4,7,BB,AC,CA
B=2,5,8,AA,BC,CB
C=0,3,6,9,AB,BA,CC
Notice that the above language uses all 9 possible ABC pairings. It will always end at either A,B,or C, and the fact that every variable use is paired means that each iteration of processing will shorten the string of variables.
Example:
1490 = AACC = BCC = BC = B (Fail)
1491 = AACA = BCA = BA = C (Success)
Not a full solution, just an idea:
(B) alone: The "plus" signs don't matter here. abc + def is the same as abcdef for the sake of divisibility by 3. For the latter case, there is a regexp here: http://blog.vkistudios.com/index.cfm/2008/12/30/Regular-Expression-to-determine-if-a-base-10-number-is-divisible-by-3
to combine this with requirement (A), we can take the solution of (B) and modify it:
First read character must be in 0..9 (not a plus)
Input must not end with a plus, so: Duplicate each state (will use S for the original state and S' for the duplicate to distinguish between them). If we're in state S and we read a plus we'll move to S'.
When reading a number we'll go to the new state as if we were in S. S' states cannot accept (another) plus.
Also, S' is not "accept state" even if S is. (because input must not end with a plus).