I was just wondering if there is a special way of saying when something equals something. For example in python, if you declare something equals 2, you say something = 2, whereas when you check if something equals something else, you would say:
if something == somethingelse:
So my question is in pseudocode for algorithms if I'm checking to see if a entered password equals a stored password in an IF THEN ELSE ENDIF loop, would I use one or two equal signs:
WHILE attempts < 3
Get EnteredPassword
**IF EnteredPassword = StoredPassword THEN**
Validated = TRUE
ELSE
attempts = attempts + 1
ENDIF
ENDWHILE
Usually, pseudocode is very broad and every author has their own way of expressing it. As
Aziz has noted, usually x <- 1 is used for an assignment and x := x + 1 for an update. Read ':=' as 'becomes' instead of 'equals', however, they are interchangeably used. As for your question, both = and == are accepted answers, as long as it is clear to your reader what your intention is.
To express equals you use the equal mark symbol once, unlike in python where you use the symbol twice to compare two values (eg if variable == 'one'). An example syntax is:
variable = 'one'
WHILE variable = 'one' DO
SEND "hi" TO DISPLAY
I'm following an exercice on exercism.io
Pretty new in the Elixir community and language, I'm doing the Elixir path to train.
Anyway, I'm hitting a brickwall and I can't understand why.
The exercice is the following :
Bob is a lackadaisical teenager. In conversation, his responses are very limited.
Bob answers 'Sure.' if you ask him a question.
He answers 'Whoa, chill out!' if you yell at him.
He answers 'Calm down, I know what I'm doing!' if you yell a question at him.
He says 'Fine. Be that way!' if you address him without actually saying anything.
He answers 'Whatever.' to anything else.
Bob's conversational partner is a purist when it comes to written communication and always follows normal rules regarding sentence punctuation in English.
So my code looks like this :
defmodule Bob do
def hey(input) do
input = String.trim(input)
is_empty? = &(String.length(&1) == 0)
is_upcase? = &(&1 == String.upcase(&1))
is_question? = &(String.ends_with?(&1, "?"))
cond do
is_question?.(input) && is_upcase?.(input) ->
"Calm down, I know what I'm doing!"
is_question?.(input) ->
"Sure."
is_empty?.(input) ->
"Fine. Be that way!"
is_upcase?.(input) == true ->
"Whoa, chill out!"
true ->
"Whatever."
end
end
end
To verify if the exercice is correct, we've got a test suite which is pretty good.
Two tests doesn't pass :
1) test only numbers (BobTest)
bob_test.exs:71
Assertion with == failed
code: assert Bob.hey("1, 2, 3") == "Whatever."
left: "Whoa, chill out!"
right: "Whatever."
stacktrace:
bob_test.exs:72: (test)
......
2) test question with numbers (BobTest)
bob_test.exs:76
Assertion with == failed
code: assert Bob.hey("4?") == "Sure."
left: "Calm down, I know what I'm doing!"
right: "Sure."
stacktrace:
bob_test.exs:77: (test)
.......
Finished in 0.09 seconds (0.08s on load, 0.01s on tests)
15 tests, 2 failures
Randomized with seed 683339
So my questions is the following : Why are numbers interpreted as upcase letters even if I pass it through String.downcase("foo") ?
String.upcase/2 returns a string with the letters converted to upper-case. There is no concept of upper and lower-case for numbers.
Your code &1 == String.upcase(&1) is comparing a string to an upper-case version of itself. In the case of a string containing no letters, those are identical. For example "1" == String.upcase("1").
The bug is that your is_upcase? function isn't actually checking if the string contains uppercase characters. As mentioned in the comments and in this answer you can add a regular expression to check that the string does actually contain an upper-case character.
By the way, the convention when using functions that end with a ? is not to prefix them with is_. So your function would be better named upcase?.
upcase? = &(&1 == String.upcase(&1) && String.match?(&1, ~r/\p{Lu}/u))
It's better to explicitly check for upper-case characters, because there are over 100,000 unicode "letters", but less than 2,000 that are considered upper-case (case-agnostic example for your tests: "聞いてる?").
I have the following code:
print("Hi there!")
def inputyournumber():
while True:
num = input("your number is?:")
if int(num) <100:
break
return
The output is:
Hi there!
your number is?: 101
your number is?: 1002
your number is?: 100
your number is?: 99
i just want the initial prints: and the final output until a correct input is entered:
Hi there!
your number is?: 99
Erasing the initially wrong inputs, but keeping the prints prior the loop. Is it possible? Or do i have to reprint them? My issue in just clearing everything then reprint the prior texts together with the correct input is that it may consume more time for more complex part of the code with similar problem.
Do you mean it only print out the number you entered when the number you enter is 100?
First, the input you get, will be a string, so the comparison won't work.
maybe this is what you wanted?
def printnum():
while True:
print('Hi there!')
num = input("your number is: ")
if num != '100':
continue
break
print(num)
printnum()
I'm using a for loop, and inside the for loop are two if-statements that are both being checked for each iteration of the for loop. My issue is that once one of the if-statements = True, then I want the True if-statement to stop being checked, and only check the second if-statement. Or vice versa.
print("Guess the numbers it is! Your goal is to guess a two digit number from 00-99. The two digits may be the same.")
numGuessesN = int(input("How many guesses would you like? (2, 3, or 4) "))
for numGuessesN in range (firstGuess, numGuessesN + 1) :
print("Guess number", numGuessesN)
userGuessN = int(input("What is your guess? "))
if (userGuessN == randomNum1) :
print("You've guessed the first number!")
break
else :
print("The first number is not", userGuessN)
if (userGuessN == randomNum2) :
print("You've guessed the second number!")
break
else :
print("The second number is not", userGuessN)
I know the breaks in the if-statements will completely exit the for loop, and that's not what I want. But somehow I need the for loop to "break" for only each individual statement if it turns out the statement is true, and keep checking the remaining statement until it's true, or the loop runs out of iterations.
Thanks guys! I'm new to all of this, so I'm sorry if I'm not really clear haha.
either you use nested if-statements, or (in case you have a lot, so you would have a lot of repeated code) you set a variable
first_wrong=False
if bla:
pass
else:
first_wrong=True
if bla2 and not first_wrong:
pass
if bla3 and not first_wrong:
pass
This question already has answers here:
Why does python use 'else' after for and while loops?
(24 answers)
Closed 7 months ago.
I have hardly ever noticed a python program that uses else in a for loop.
I recently used it to perform an action based on the loop variable condition while exiting; as it is in the scope.
What is the pythonic way to use an else in a for loop? Are there any notable use cases?
And, yea. I dislike using break statement. I'd rather set the looping condition complex. Would I be able to get any benefit out of it, if I don't like to use break statement anyway.
Worth noting that for loop has an else since the language inception, the first ever version.
What could be more pythonic than PyPy?
Look at what I discovered starting at line 284 in ctypes_configure/configure.py:
for i in range(0, info['size'] - csize + 1, info['align']):
if layout[i:i+csize] == [None] * csize:
layout_addfield(layout, i, ctype, '_alignment')
break
else:
raise AssertionError("unenforceable alignment %d" % (
info['align'],))
And here, from line 425 in pypy/annotation/annrpython.py (clicky)
if cell.is_constant():
return Constant(cell.const)
else:
for v in known_variables:
if self.bindings[v] is cell:
return v
else:
raise CannotSimplify
In pypy/annotation/binaryop.py, starting at line 751:
def is_((pbc1, pbc2)):
thistype = pairtype(SomePBC, SomePBC)
s = super(thistype, pair(pbc1, pbc2)).is_()
if not s.is_constant():
if not pbc1.can_be_None or not pbc2.can_be_None:
for desc in pbc1.descriptions:
if desc in pbc2.descriptions:
break
else:
s.const = False # no common desc in the two sets
return s
A non-one-liner in pypy/annotation/classdef.py, starting at line 176:
def add_source_for_attribute(self, attr, source):
"""Adds information about a constant source for an attribute.
"""
for cdef in self.getmro():
if attr in cdef.attrs:
# the Attribute() exists already for this class (or a parent)
attrdef = cdef.attrs[attr]
s_prev_value = attrdef.s_value
attrdef.add_constant_source(self, source)
# we should reflow from all the reader's position,
# but as an optimization we try to see if the attribute
# has really been generalized
if attrdef.s_value != s_prev_value:
attrdef.mutated(cdef) # reflow from all read positions
return
else:
# remember the source in self.attr_sources
sources = self.attr_sources.setdefault(attr, [])
sources.append(source)
# register the source in any Attribute found in subclasses,
# to restore invariant (III)
# NB. add_constant_source() may discover new subdefs but the
# right thing will happen to them because self.attr_sources
# was already updated
if not source.instance_level:
for subdef in self.getallsubdefs():
if attr in subdef.attrs:
attrdef = subdef.attrs[attr]
s_prev_value = attrdef.s_value
attrdef.add_constant_source(self, source)
if attrdef.s_value != s_prev_value:
attrdef.mutated(subdef) # reflow from all read positions
Later in the same file, starting at line 307, an example with an illuminating comment:
def generalize_attr(self, attr, s_value=None):
# if the attribute exists in a superclass, generalize there,
# as imposed by invariant (I)
for clsdef in self.getmro():
if attr in clsdef.attrs:
clsdef._generalize_attr(attr, s_value)
break
else:
self._generalize_attr(attr, s_value)
If you have a for loop you don't really have any condition statement. So break is your choice if you like to abort and then else can serve perfectly to handle the case where you were not happy.
for fruit in basket:
if fruit.kind in ['Orange', 'Apple']:
fruit.eat()
break
else:
print 'The basket contains no desirable fruit'
Basically, it simplifies any loop that uses a boolean flag like this:
found = False # <-- initialize boolean
for divisor in range(2, n):
if n % divisor == 0:
found = True # <-- update boolean
break # optional, but continuing would be a waste of time
if found: # <-- check boolean
print n, "is composite"
else:
print n, "is prime"
and allows you to skip the management of the flag:
for divisor in range(2, n):
if n % divisor == 0:
print n, "is composite"
break
else:
print n, "is prime"
Note that there is already a natural place for code to execute when you do find a divisor - right before the break. The only new feature here is a place for code to execute when you tried all divisor and did not find any.
This helps only in conjuction with break. You still need booleans if you can't break (e.g. because you looking for the last match, or have to track several conditions in parallel).
Oh, and BTW, this works for while loops just as well.
any/all
Nowdays, if the only purpose of the loop is a yes-or-no answer, you might be able to write it much shorter with the any()/all() functions with a generator or generator expression that yields booleans:
if any(n % divisor == 0
for divisor in range(2, n)):
print n, "is composite"
else:
print n, "is prime"
Note the elegancy! The code is 1:1 what you want to say!
[This is as effecient as a loop with a break, because the any() function is short-circuiting, only running the generator expression until it yeilds True. In fact it's usually even faster than a loop. Simpler Python code tends to have less overhear.]
This is less workable if you have other side effects - for example if you want to find the divisor. You can still do it (ab)using the fact that non-0 value are true in Python:
divisor = any(d for d in range(2, n) if n % d == 0)
if divisor:
print n, "is divisible by", divisor
else:
print n, "is prime"
but as you see this is getting shaky - wouldn't work if 0 was a possible divisor value...
Without using break, else blocks have no benefit for for and while statements. The following two examples are equivalent:
for x in range(10):
pass
else:
print "else"
for x in range(10):
pass
print "else"
The only reason for using else with for or while is to do something after the loop if it terminated normally, meaning without an explicit break.
After a lot of thinking, I can finally come up with a case where this might be useful:
def commit_changes(directory):
for file in directory:
if file_is_modified(file):
break
else:
# No changes
return False
# Something has been changed
send_directory_to_server()
return True
Perhaps the best answer comes from the official Python tutorial:
break and continue Statements, and else Clauses on Loops:
Loop statements may have an else
clause; it is executed when the loop
terminates through exhaustion of the
list (with for) or when the condition
becomes false (with while), but not
when the loop is terminated by a break
statement
I was introduced to a wonderful idiom in which you can use a for/break/else scheme with an iterator to save both time and LOC. The example at hand was searching for the candidate for an incompletely qualified path. If you care to see the original context, please see the original question.
def match(path, actual):
path = path.strip('/').split('/')
actual = iter(actual.strip('/').split('/'))
for pathitem in path:
for item in actual:
if pathitem == item:
break
else:
return False
return True
What makes the use of for/else so great here is the elegance of avoiding juggling a confusing boolean around. Without else, but hoping to achieve the same amount of short-circuiting, it might be written like so:
def match(path, actual):
path = path.strip('/').split('/')
actual = iter(actual.strip('/').split('/'))
failed = True
for pathitem in path:
failed = True
for item in actual:
if pathitem == item:
failed = False
break
if failed:
break
return not failed
I think the use of else makes it more elegant and more obvious.
A use case of the else clause of loops is breaking out of nested loops:
while True:
for item in iterable:
if condition:
break
suite
else:
continue
break
It avoids repeating conditions:
while not condition:
for item in iterable:
if condition:
break
suite
Here you go:
a = ('y','a','y')
for x in a:
print x,
else:
print '!'
It's for the caboose.
edit:
# What happens if we add the ! to a list?
def side_effect(your_list):
your_list.extend('!')
for x in your_list:
print x,
claimant = ['A',' ','g','u','r','u']
side_effect(claimant)
print claimant[-1]
# oh no, claimant now ends with a '!'
edit:
a = (("this","is"),("a","contrived","example"),("of","the","caboose","idiom"))
for b in a:
for c in b:
print c,
if "is" == c:
break
else:
print