Related
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
Can you wrap a query function with an if function?
My query can produce 3 results, yes yes, yes no, no no. These results occur in two different cells. I want the if function to then say - yes yes - true and either of the other results false. When doing it simply it just gives an error.
Wrapping of formulas
The IF formula has the syntax IF(logical_expression, value_if_true, value_if_false)
Thereby, the logical expression can be a comparison of values, e.g IF(value="Yes", TRUE, FALSE)
value can be in this case either
a fixed value, e.g. a string
a cell reference, e.g. IF(J2="Yes", TRUE, FALSE)
a nested formula, which can a query
In your case, you are interested in the latter, so:
=IF(query(Sheet2!A1:J7,"select G where A = '"&A2&"' and '"&E2&"' = B",0)="Yes", TRUE, FALSE)
I was able to get an answer with using the IFS function =ifs(query(Sheet2!A3:J9,"select G where A = '"&A4&"' and '"&E4&"' = B",0)="yes","TRUE",query(Sheet2!A3:J9,"select G where A = '"&A4&"' and '"&E4&"' = B",0)="no","FALSE")
That was able to solve the issue that the simple if function could not solve
I want to evaluate if a constraint is respected or not in Pyomo when the values of the variables contained in constraint expression are known.
Use case: We know that one particular constraint sometimes makes the problem infeasible, depending on the value of the variable. Instead of sending the problem to the solver to test if the problem is feasible, converting the constraint expression to a boolean type would be enough to determine if the constraint is the culprit.
For the sake of providing a feasible example, here would be the code:
from pyomo.environ import ConcreteModel, Var, Constraint
model = ConcreteModel()
model.X = Var()
def constraint_rule(model):
return model.X <= 1
model.a_constraint = Constraint(rule=constraint_rule)
Now, let's try to work with the expression to evaluate:
# Let's define the expression in this way:
expression = constraint_rule(model)
# Let's show that the expression is what we expected:
print(str(expression))
The previous statement should print X <= 1.0.
Now, the tricky part is how to evaluate the expression.
if expression == True:
print("feasible")
else:
print("infeasible")
creates an TypeError Exception (TypeError: Cannot create an EqualityExpression where one of the sub-expressions is a relational expression: X <= 1.0).
The last example doesn't work because constraint_rule doesn't return a boolean but a Pyomo expression.
Finally, I know that something like
def evaluate_constraint_a_expression(model):
return value(model.X) <= 1
would work, but I can't assume that I will always know the content of my constraint expression, so I need a robust way of evaluating it.
Is there a clever way of achieving this? Like, evaluating the expression as a boolean and evaluating the left hand side and right hand side of the expression at the same time?
The solution is to use value function. Even if it says that it evaluates an expression to a numeric value, it also converts the expression to a boolean value if it is an equality/inequality expression, like the rule of a constraint.
Let's suppose that the model is defined the way it is in the question, then the rest of the code should be:
from pyomo.environ import value
if value(expression) == True:
print("feasible")
else:
print("infeasible")
where expression is defined as written in the question.
However, be advised that numeric precision in Python using this method can be different than the one provided by the solver. Therefore, it is possible that this method will show that a constraint is infeasible while it is just a matter of numeric imprecision of under 1e-10. So, while it is useful in finding if most constraints are feasible, it also generates some false positives.
Let's suppose that the objective function is
max z(x,y) = f1(x) - f2(y)
where f1 is function of variables x and f2 is functions of variables y.
This could be written in Pyomo as
def z(model):
return f1(model) - f2(model)
def f1(model):
return [some summation of x variables with some coefficients]
def f2(model):
return [some summation of y variables with some coefficients]
model.objective = Objective(rule=z)
I know it is possible to get the numeric value of z(x,y) easily by calling (since it is the objective function) :
print(model.objective())
but is there a way to get the numeric value of any of these sub-functions separetedly after the optimization, even if they are not explicitly defined as objectives?
I'll answer your question in terms of a ConcreteModel, since rules in Pyomo, for the most part, are nothing more than a mechanism to delay building a ConcereteModel. For now, they are also required to define indexed objects, but that will likely change soon.
First, there is nothing stopping you from defining those "rules" as standard functions that take in some argument and return a value. E.g.,
def z(x, y):
return f1(x) - f2(y)
def f1(x):
return x + 1
def f2(x):
return y**2
Now if you call any of these functions with a built-in type (e.g., f(1,5)), you will get a number back. However, if you call them with Pyomo variables (or Pyomo expressions) you will get a Pyomo expression back, which you can assign to an objective or constraint. This works because Pyomo modeling components, such as variables, overload the standard algebraic operators like +, -, *, etc. Here is an example of how you can build an objective with these functions:
import pyomo.environ as aml
m = aml.ConcreteModel()
m.x = aml.Var()
m.y = aml.Var()
m.o = aml.Objective(expr= z(m.x, m.y))
Now if m.x and m.y have a value loaded into them (i.e., the .value attribute is something other than None), then you can call one of the sub-functions with them and evaluate the returned expression (slower)
aml.value(f1(m.x))
aml.value(f2(m.y))
or you can extract the value from them and pass that to the sub-functions (faster)
f1(m.x.value)
f2(m.y.value)
You can also use the Expression object to store sub-expressions that you want to evaluate on the fly or share inside multiple other expression on a model (all of which you can update by changing what expression is stored under the Expression object).
This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
My Google-fu has failed me.
In Python, are the following two tests for equality equivalent?
n = 5
# Test one.
if n == 5:
print 'Yay!'
# Test two.
if n is 5:
print 'Yay!'
Does this hold true for objects where you would be comparing instances (a list say)?
Okay, so this kind of answers my question:
L = []
L.append(1)
if L == [1]:
print 'Yay!'
# Holds true, but...
if L is [1]:
print 'Yay!'
# Doesn't.
So == tests value where is tests to see if they are the same object?
is will return True if two variables point to the same object (in memory), == if the objects referred to by the variables are equal.
>>> a = [1, 2, 3]
>>> b = a
>>> b is a
True
>>> b == a
True
# Make a new copy of list `a` via the slice operator,
# and assign it to variable `b`
>>> b = a[:]
>>> b is a
False
>>> b == a
True
In your case, the second test only works because Python caches small integer objects, which is an implementation detail. For larger integers, this does not work:
>>> 1000 is 10**3
False
>>> 1000 == 10**3
True
The same holds true for string literals:
>>> "a" is "a"
True
>>> "aa" is "a" * 2
True
>>> x = "a"
>>> "aa" is x * 2
False
>>> "aa" is intern(x*2)
True
Please see this question as well.
There is a simple rule of thumb to tell you when to use == or is.
== is for value equality. Use it when you would like to know if two objects have the same value.
is is for reference equality. Use it when you would like to know if two references refer to the same object.
In general, when you are comparing something to a simple type, you are usually checking for value equality, so you should use ==. For example, the intention of your example is probably to check whether x has a value equal to 2 (==), not whether x is literally referring to the same object as 2.
Something else to note: because of the way the CPython reference implementation works, you'll get unexpected and inconsistent results if you mistakenly use is to compare for reference equality on integers:
>>> a = 500
>>> b = 500
>>> a == b
True
>>> a is b
False
That's pretty much what we expected: a and b have the same value, but are distinct entities. But what about this?
>>> c = 200
>>> d = 200
>>> c == d
True
>>> c is d
True
This is inconsistent with the earlier result. What's going on here? It turns out the reference implementation of Python caches integer objects in the range -5..256 as singleton instances for performance reasons. Here's an example demonstrating this:
>>> for i in range(250, 260): a = i; print "%i: %s" % (i, a is int(str(i)));
...
250: True
251: True
252: True
253: True
254: True
255: True
256: True
257: False
258: False
259: False
This is another obvious reason not to use is: the behavior is left up to implementations when you're erroneously using it for value equality.
Is there a difference between == and is in Python?
Yes, they have a very important difference.
==: check for equality - the semantics are that equivalent objects (that aren't necessarily the same object) will test as equal. As the documentation says:
The operators <, >, ==, >=, <=, and != compare the values of two objects.
is: check for identity - the semantics are that the object (as held in memory) is the object. Again, the documentation says:
The operators is and is not test for object identity: x is y is true
if and only if x and y are the same object. Object identity is
determined using the id() function. x is not y yields the inverse
truth value.
Thus, the check for identity is the same as checking for the equality of the IDs of the objects. That is,
a is b
is the same as:
id(a) == id(b)
where id is the builtin function that returns an integer that "is guaranteed to be unique among simultaneously existing objects" (see help(id)) and where a and b are any arbitrary objects.
Other Usage Directions
You should use these comparisons for their semantics. Use is to check identity and == to check equality.
So in general, we use is to check for identity. This is usually useful when we are checking for an object that should only exist once in memory, referred to as a "singleton" in the documentation.
Use cases for is include:
None
enum values (when using Enums from the enum module)
usually modules
usually class objects resulting from class definitions
usually function objects resulting from function definitions
anything else that should only exist once in memory (all singletons, generally)
a specific object that you want by identity
Usual use cases for == include:
numbers, including integers
strings
lists
sets
dictionaries
custom mutable objects
other builtin immutable objects, in most cases
The general use case, again, for ==, is the object you want may not be the same object, instead it may be an equivalent one
PEP 8 directions
PEP 8, the official Python style guide for the standard library also mentions two use-cases for is:
Comparisons to singletons like None should always be done with is or
is not, never the equality operators.
Also, beware of writing if x when you really mean if x is not None --
e.g. when testing whether a variable or argument that defaults to None
was set to some other value. The other value might have a type (such
as a container) that could be false in a boolean context!
Inferring equality from identity
If is is true, equality can usually be inferred - logically, if an object is itself, then it should test as equivalent to itself.
In most cases this logic is true, but it relies on the implementation of the __eq__ special method. As the docs say,
The default behavior for equality comparison (== and !=) is based on
the identity of the objects. Hence, equality comparison of instances
with the same identity results in equality, and equality comparison of
instances with different identities results in inequality. A
motivation for this default behavior is the desire that all objects
should be reflexive (i.e. x is y implies x == y).
and in the interests of consistency, recommends:
Equality comparison should be reflexive. In other words, identical
objects should compare equal:
x is y implies x == y
We can see that this is the default behavior for custom objects:
>>> class Object(object): pass
>>> obj = Object()
>>> obj2 = Object()
>>> obj == obj, obj is obj
(True, True)
>>> obj == obj2, obj is obj2
(False, False)
The contrapositive is also usually true - if somethings test as not equal, you can usually infer that they are not the same object.
Since tests for equality can be customized, this inference does not always hold true for all types.
An exception
A notable exception is nan - it always tests as not equal to itself:
>>> nan = float('nan')
>>> nan
nan
>>> nan is nan
True
>>> nan == nan # !!!!!
False
Checking for identity can be much a much quicker check than checking for equality (which might require recursively checking members).
But it cannot be substituted for equality where you may find more than one object as equivalent.
Note that comparing equality of lists and tuples will assume that identity of objects are equal (because this is a fast check). This can create contradictions if the logic is inconsistent - as it is for nan:
>>> [nan] == [nan]
True
>>> (nan,) == (nan,)
True
A Cautionary Tale:
The question is attempting to use is to compare integers. You shouldn't assume that an instance of an integer is the same instance as one obtained by another reference. This story explains why.
A commenter had code that relied on the fact that small integers (-5 to 256 inclusive) are singletons in Python, instead of checking for equality.
Wow, this can lead to some insidious bugs. I had some code that checked if a is b, which worked as I wanted because a and b are typically small numbers. The bug only happened today, after six months in production, because a and b were finally large enough to not be cached. – gwg
It worked in development. It may have passed some unittests.
And it worked in production - until the code checked for an integer larger than 256, at which point it failed in production.
This is a production failure that could have been caught in code review or possibly with a style-checker.
Let me emphasize: do not use is to compare integers.
== determines if the values are equal, while is determines if they are the exact same object.
What's the difference between is and ==?
== and is are different comparison! As others already said:
== compares the values of the objects.
is compares the references of the objects.
In Python names refer to objects, for example in this case value1 and value2 refer to an int instance storing the value 1000:
value1 = 1000
value2 = value1
Because value2 refers to the same object is and == will give True:
>>> value1 == value2
True
>>> value1 is value2
True
In the following example the names value1 and value2 refer to different int instances, even if both store the same integer:
>>> value1 = 1000
>>> value2 = 1000
Because the same value (integer) is stored == will be True, that's why it's often called "value comparison". However is will return False because these are different objects:
>>> value1 == value2
True
>>> value1 is value2
False
When to use which?
Generally is is a much faster comparison. That's why CPython caches (or maybe reuses would be the better term) certain objects like small integers, some strings, etc. But this should be treated as implementation detail that could (even if unlikely) change at any point without warning.
You should only use is if you:
want to check if two objects are really the same object (not just the same "value"). One example can be if you use a singleton object as constant.
want to compare a value to a Python constant. The constants in Python are:
None
True1
False1
NotImplemented
Ellipsis
__debug__
classes (for example int is int or int is float)
there could be additional constants in built-in modules or 3rd party modules. For example np.ma.masked from the NumPy module)
In every other case you should use == to check for equality.
Can I customize the behavior?
There is some aspect to == that hasn't been mentioned already in the other answers: It's part of Pythons "Data model". That means its behavior can be customized using the __eq__ method. For example:
class MyClass(object):
def __init__(self, val):
self._value = val
def __eq__(self, other):
print('__eq__ method called')
try:
return self._value == other._value
except AttributeError:
raise TypeError('Cannot compare {0} to objects of type {1}'
.format(type(self), type(other)))
This is just an artificial example to illustrate that the method is really called:
>>> MyClass(10) == MyClass(10)
__eq__ method called
True
Note that by default (if no other implementation of __eq__ can be found in the class or the superclasses) __eq__ uses is:
class AClass(object):
def __init__(self, value):
self._value = value
>>> a = AClass(10)
>>> b = AClass(10)
>>> a == b
False
>>> a == a
So it's actually important to implement __eq__ if you want "more" than just reference-comparison for custom classes!
On the other hand you cannot customize is checks. It will always compare just if you have the same reference.
Will these comparisons always return a boolean?
Because __eq__ can be re-implemented or overridden, it's not limited to return True or False. It could return anything (but in most cases it should return a boolean!).
For example with NumPy arrays the == will return an array:
>>> import numpy as np
>>> np.arange(10) == 2
array([False, False, True, False, False, False, False, False, False, False], dtype=bool)
But is checks will always return True or False!
1 As Aaron Hall mentioned in the comments:
Generally you shouldn't do any is True or is False checks because one normally uses these "checks" in a context that implicitly converts the condition to a boolean (for example in an if statement). So doing the is True comparison and the implicit boolean cast is doing more work than just doing the boolean cast - and you limit yourself to booleans (which isn't considered pythonic).
Like PEP8 mentions:
Don't compare boolean values to True or False using ==.
Yes: if greeting:
No: if greeting == True:
Worse: if greeting is True:
They are completely different. is checks for object identity, while == checks for equality (a notion that depends on the two operands' types).
It is only a lucky coincidence that "is" seems to work correctly with small integers (e.g. 5 == 4+1). That is because CPython optimizes the storage of integers in the range (-5 to 256) by making them singletons. This behavior is totally implementation-dependent and not guaranteed to be preserved under all manner of minor transformative operations.
For example, Python 3.5 also makes short strings singletons, but slicing them disrupts this behavior:
>>> "foo" + "bar" == "foobar"
True
>>> "foo" + "bar" is "foobar"
True
>>> "foo"[:] + "bar" == "foobar"
True
>>> "foo"[:] + "bar" is "foobar"
False
https://docs.python.org/library/stdtypes.html#comparisons
is tests for identity
== tests for equality
Each (small) integer value is mapped to a single value, so every 3 is identical and equal. This is an implementation detail, not part of the language spec though
Your answer is correct. The is operator compares the identity of two objects. The == operator compares the values of two objects.
An object's identity never changes once it has been created; you may think of it as the object's address in memory.
You can control comparison behaviour of object values by defining a __cmp__ method or a rich comparison method like __eq__.
Have a look at Stack Overflow question Python's “is” operator behaves unexpectedly with integers.
What it mostly boils down to is that "is" checks to see if they are the same object, not just equal to each other (the numbers below 256 are a special case).
In a nutshell, is checks whether two references point to the same object or not.== checks whether two objects have the same value or not.
a=[1,2,3]
b=a #a and b point to the same object
c=list(a) #c points to different object
if a==b:
print('#') #output:#
if a is b:
print('##') #output:##
if a==c:
print('###') #output:##
if a is c:
print('####') #no output as c and a point to different object
As the other people in this post answer the question in details the difference between == and is for comparing Objects or variables, I would emphasize mainly the comparison between is and == for strings which can give different results and I would urge programmers to carefully use them.
For string comparison, make sure to use == instead of is:
str = 'hello'
if (str is 'hello'):
print ('str is hello')
if (str == 'hello'):
print ('str == hello')
Out:
str is hello
str == hello
But in the below example == and is will get different results:
str2 = 'hello sam'
if (str2 is 'hello sam'):
print ('str2 is hello sam')
if (str2 == 'hello sam'):
print ('str2 == hello sam')
Out:
str2 == hello sam
Conclusion and Analysis:
Use is carefully to compare between strings.
Since is for comparing objects and since in Python 3+ every variable such as string interpret as an object, let's see what happened in above paragraphs.
In python there is id function that shows a unique constant of an object during its lifetime. This id is using in back-end of Python interpreter to compare two objects using is keyword.
str = 'hello'
id('hello')
> 140039832615152
id(str)
> 140039832615152
But
str2 = 'hello sam'
id('hello sam')
> 140039832615536
id(str2)
> 140039832615792
As John Feminella said, most of the time you will use == and != because your objective is to compare values. I'd just like to categorise what you would do the rest of the time:
There is one and only one instance of NoneType i.e. None is a singleton. Consequently foo == None and foo is None mean the same. However the is test is faster and the Pythonic convention is to use foo is None.
If you are doing some introspection or mucking about with garbage collection or checking whether your custom-built string interning gadget is working or suchlike, then you probably have a use-case for foo is bar.
True and False are also (now) singletons, but there is no use-case for foo == True and no use case for foo is True.
Most of them already answered to the point. Just as an additional note (based on my understanding and experimenting but not from a documented source), the statement
== if the objects referred to by the variables are equal
from above answers should be read as
== if the objects referred to by the variables are equal and objects belonging to the same type/class
. I arrived at this conclusion based on the below test:
list1 = [1,2,3,4]
tuple1 = (1,2,3,4)
print(list1)
print(tuple1)
print(id(list1))
print(id(tuple1))
print(list1 == tuple1)
print(list1 is tuple1)
Here the contents of the list and tuple are same but the type/class are different.