Is using If, ElseIf, ElseIf better than using If, If, If? - if-statement

Is there really any difference between using
If(this)
{
}
Else If(that)
{
}
Else
{
}
or using,
If(this)
{
}
If(that)
{
}
Else
{
}
?
Does one execute any faster? Does the compiler or architecture make any difference?


There's the huge difference that the contents of the this-block and the that-block can both be executed in the second form, whereas the first form allows at most one of those to be executed.
Compare these two Python snippets:
x = 10
if x > 5:
print "x larger than 5"
elif x > 1:
print "x larger than 1"
else:
print "x not larger than 1"
# output: x larger than 5
and
x = 10
if x > 5:
print "x larger than 5"
if x > 1: # not using else-if anymore!
print "x larger than 1"
else:
print "x not larger than 1"
# output line 1: x larger than 5
# output line 2: x larger than 1
As others have mentioned, you generally shouldn't be concerned about performance between these variations so much as you should be concerned about correctness. However, since you asked... all else being equal, the second form will be slower because the second conditional must be evaluated.
But unless you have determined that code written in this form is a bottleneck, it's not really worth your effort to even think about optimizing it. In switching from the first form to the second, you give up the jump out of the first if-statement and get back a forced evaluation of the second condition. Depending on the language, that's probably an extremely negligible difference.


Yes, in your first example, if this evaluates to true and that evaluates to true, only the first code block will be executed, whereas in the second example, they both will.
They are not equivalent


Yes.
In the First case: control-flow will only check the next condition if the current condition fails but in the second case it will check all conditions that come across.
In first case Else part will only be executed if all previous conditions fails to be true. and in the second case only if the last If condition fails.


it makes a difference.
in case "this" and "that" are both true, both part of code will result something else.


In first code.
it will check IF, if this is true then execute its body,
if false then will
check ELSEIF if that is true that execute its body,
if false then will
execute body of else.
Second code
it will check IF, if this is true then execute its body,
if false do nothing.
check 2nd IF if that is true that execute its body,
if false then will
execute body of else.

Related

Check if all elements of list are prime in Raku

my #g = (1,2,3,4);
say reduce {is-prime}, #g; # ==> gives error
say reduce {is-prime *}, #g; #==> gives error
say reduce {is-prime}, (1,2,3,4); # ==> gives error
say so is-prime #g.all; # ==> gives error
How to check if all elements of list are prime in Raku?

The answers above are all helpful, but they fail to explain why your solution does not work. Basically reduce is not going to apply a function (in your case, is-prime) to every member of a list. You want map for that. The error says
Calling is-prime() will never work with signature of the proto ($, *%)
Because reduce expects an infix, thus binary, function, or a function with two arguments; what it does is to apply them to the first pair of elements, then to the result and the third element, and so on. Last statement does not work for a similar reason: you are calling is-prime with a list argument, not a single argument.

You're basically asking: are there any elements in this list which are not prime? I would write that as:
say "not all prime" if #g.first: !*.is-prime;
Please note though, that apparently 1 is not considered prime according to the is-prime function:
say 1.is-prime; # False
so the first would trigger on the 1 in your example, not on the 4.

There are of course may ways to do this. A very explicit way is using a for loop:
for #g -> $g {
if $g.is-prime {
say $g;
}
}
Or with a grep (you could leave the $_ implicit):
#g.grep({ $_.is-prime }).say
Both above are assuming you really want to filter the primes out. Of course you can also really check each number and get a boolean:
#g.map({ .is-prime }).say

There is a big problem with this:
say reduce {is-prime}, #g;
You created a lambda:
{ }
The only thing it does is calls a function:
is-prime
You didn't give the function any arguments though.
Is it just supposed to guess what the arguments should be?
If you meant to pass in is-prime as a reference, you should have used &is-prime rather than {is-prime}.
Of course that still wouldn't have worked.
The other problem is that reduce operates by recursively combining values.
It can't do that if it operates on one argument at a time.
The bare block lambda {}, takes zero or one argument, not two or more.
reduce is often combined with map.
It happens so often that there is a Wikipedia page about MapReduce.
say ( map &is-prime, #g ==> reduce { $^a and $^b } );
# False
say ( map &is-prime, 2,3,5 ==> reduce { $^a and $^b } );
# True
I wrote it that way so that map would be in the line before reduce, but perhaps it would be more clear this way:
say reduce {$^a and $^b}, map &is-prime, 2,3,5;
# True
reduce with an infix operator is so common that there is a shorter way to write it.
say [and] map &is-prime, 2,3,5;
# True
Of course it would be better to just find the first value that isn't prime, and say the inverse.
Since if there is even a single value that isn't prime that would mean they can't all be primes.
You have to be careful though, as you may think something like this would always work:
not #g.first: !*.is-prime;
It does happen to work for the values you gave it, but may not always.
first returns Nil if it can't find the value.
not (2,3,5).first: !*.is-prime;
# not Nil === True
not (2,3,4).first: !*.is-prime;
# not 4 === False
not (2,3,0,4).first: !*.is-prime;
# not 0 === True
That last one returned 0 which when combined with not returns True.
You could fix this with defined.
not defined (2,3,0,4).first: !*.is-prime;
# False
This only works if first wouldn't return an undefined element that happens to be in the list.
(Int,Any).first: Real
# Int
defined (Int,Any).first: Real
# False
You could fix that by asking for the index instead of the value.
You of course still need defined.
(Int,Any).first: :k, Real
# 0
defined (Int,Any).first: :k, Real
# True
The other way to fix it is to just use grep.
not (2,3,0,4).grep: !*.is-prime;
# not (0,4) === False
Since grep always returns a List, you don't have to worry about checking for 0 or undefined elements.
(A List is True if it contains any elements, no matter what the values.)
grep is smart enough to know that if you coerce to Bool that it can stop upon finding the first value.
So it short-circuits the same as if you had used first.
This results in some fairly funky code, with those two negating operators. So it should be put into a function.
sub all-prime ( +#_ ) {
# return False if we find any non-prime
not #_.grep: !*.is-prime
# grep short-circuits in Bool context, so this will stop early
}
This could still fail if you give it something weird
all-prime 2,3,5, Date.today;
# ERROR: No such method 'is-prime' for invocant of type 'Date'
If you care, add some error handling.
sub all-prime ( +#_ ) {
# return Nil if there was an error
CATCH { default { return Nil }}
# return False if we find any non-prime
not #_.grep: !*.is-prime
}
all-prime 2,3,5, Date.today;
# Nil

use the all junction:
say so all #g».is-prime; # False

Testing variable vs multiple true statements but stoping in the very first is true (If operator)

I´m starting to develop a piping calculation software, at some point it reads the input pressure and is supposed to return a value (pipe schedule) corresponding to the pressure value. The problem is that if require to get the "schedule" given the pressure it is likely my statement is true for all the possibilities, but I want it to stop and return the value when the very first statement is fulfilled. In the following example assume the pressure input is "60", so it is true for all the statements (167, 241 and 397) but I need it to stop when the very first is met, or continue with the following if the first one is not true:
diameter_= float(diameter.get())
pressure_= float(pipepressure.get())
if diameter_== 2:
if (pressure_*1.5) < 167:
schedule.set("40")
if (pressure_*1.5) < 241:
schedule.set("80")
if (pressure_*1.5) < 397:
schedule.set("160")

How to fix a loop with boolean variable

I'm doing a project for a class, and I opted to make a text based game in python. I'm trying to set it up so that the question will loop until the player confirms their choice, and I'm having problems with the while loop in this section.
def pc_cls_sc(x):
# code does some stuff
print "You are sure about" + str(x)
exVar = raw_input("Right?")
if exVar == "y":
print "Alright!"
conf_Class = True
else:
print "Ok then."
conf_Class = False
while conf_Class is False:
pc_Class = raw_input(#asks some question)
pc_cls_sc(pc_Class)
The rest of this code functions properly, but the loop continues after the conf_Class variable is supposed to be set to true. I have a similar loop earlier in my code, which works just fine. I've tried moving the variable reassignment outside of the pc_cls_sc function, but all it did was cause double output. Can anyone tell me how to fix this?

You can use break to exit the loop. The code below will keep asking a user for input until they say 'y'.
while True:
x=input("Right? " )
if (x=='y'):
break

Order of if statements?

I wonder if the first if statement is True, whether are the following elif executed or not (even if they are also True) ?
I have tried:
if True:
print "Hi"
elif True:
print "hello"
And it prints only "Hi". But in a more complex part of my code, I see some print outputs displayed that are within some elif statements and I just basically switch the very first if to True so that none of the elif should be executed for testing purposes. So why are the print statement that are within the elif printed?
(code is too long to print here, I just wonder if it could be an answer without...)

An elif will only be entered if the condition matches and the previous condition does not:
>>> if x == True:
... print 'hi'
... elif x == False:
... print 'bye'
...
hi
>>> if x == True:
... print 'hi'
... elif x == True:
... print 'bye' #not entered because previous condition was valid
...
hi
>>>

I'm not sure if I understand this correctly, but have you tried else?
if True:
print "Hi"
else:
print "Hello"
EDIT: Sorry, understood what you mean now..
I might be wrong, but "elif" means "else if"
So what you're basicly writing is like this:
If i have 5 apples, i put up a hand
else if i have 5 apples, i put up a leg.
using elif should however not be used at the end of a statement/code. If you aren't going to compare more than 2 statements, use 'if' and 'else'
But why should you have 2 'True'?

In an [if, elif, else] blocks only the first block with the condition that checks out will be executed.
but considering your situation I guess it needs more context to understand it fully, post part of the code you're testing or something with the same logic.

How to create multiple loops in python

I'm trying to create a basic maths program which will randomly generate a numerical question and then allows 3 attempts to complete it before it moves onto the next question however cant figure out how to make it do both together.
My code currently looks like this
print "What is your name?"
score = 0
Attempt = 0
loop = True
Name = raw_input()
import random
for i in range (1,6):
question_1_part_1 = random.randint(1,30)
question_1_part_2 = random.randint(1,30)
print "What is", question_1_part_1, "+", question_1_part_2, "?"
while Attempt <3: # inputing this while loop here means it will retry the same question.
Guess = raw_input()
Guess = int(Guess)
answer = question_1_part_1 + question_1_part_2
if Guess == answer:
print "Well done"
score = score + 1
else: print "try again"
Attempt = Attempt + 1
if Attempt == 3: print "You failed"
print Name, "your score is", score

A simple break statement will take you out of the loop when the answer is correct.
if Guess == answer:
print "Well done"
score += 1
break
else: print "try again"
Note the change to your assignment; this is considered a cleaner increment.
To answer your comment ... you don't use this when the person gets the question wrong. Your program logic freezes out the user after three wrong guesses total, on all questions. If you want them to have up to three guesses on every problem, then you have to reset the Attempt counter for every question. Pull that line into the outer loop:
for i in range (1,6):
Attempt = 0
loop = True
question_1_part_1 = random.randint(1,30)
question_1_part_2 = random.randint(1,30)
In the future, I strongly recommend that you program incrementally: get a few lines working before you add more. Insert tracing print statements to check that the values and program flow are correct. Don't remove them until that part of the code is solid -- and even then only comment them out until the entire program is working. Some of your problems stem from trying to write more than you can comfortably hold in your mind at once -- which is common at most levels of programming. :-)

Add another argument to your while loop.
while Attempt <3 and Guess != ... :
rest of loop code
Then you are exiting your loop when they get the correct answer. Or you could break from the statement when they get the answer right.