Please explain "set difference" in python - python

Trying to learn Python I encountered the following:
>>> set('spam') - set('ham')
set(['p', 's'])
Why is it set(['p', 's']) - i mean: why is 'h' missing?

The - operator on python sets is mapped to the difference method, which is defined as the members of set A which are not members of set B. So in this case, the members of "spam" which are not in "ham"are "s" and "p". Notice that this method is not commutative (that is, a - b == b - a is not always true).
You may be looking for the symmetric_difference or ^ method:
>>> set("spam") ^ set("ham")
{'h', 'p', 's'}
This operator is commutative.

Because that is the definition of a set difference. In plain English, it is equivalent to "what elements are in A that are not also in B?".
Note the reverse behavior makes this more obvious
>>> set('spam') - set('ham')
{'s', 'p'}
>>> set('ham') - set('spam')
{'h'}
To get all unique elements, disregarding the order in which you ask, you can use symmetric_difference
>>> set('spam').symmetric_difference(set('ham'))
{'s', 'h', 'p'}

There are two different operators:
Set difference. This is defined as the elements of A not present in B, and is written as A - B or A.difference(B).
Symmetric set difference. This is defined as the elements of either set not present in the other set, and is written as A ^ B or A.symmetric_difference(B).
Your code is using the former, whereas you seem to be expecting the latter.

The set difference is the set of all characters in the first set that are not in the second set. 'p' and 's' appear in the first set but not in the second, so they are in the set difference. 'h' does not appear in the first set, so it is not in the set difference (regardless of whether or not it is in the first set).

You can also obtain the desired result as:
>>> (set('spam') | set('ham')) - (set('spam') & set('ham'))
set(['p', 's', 'h'])
Create union using | and intersection using & and then do the set difference, i.e. differences between all elements and common elements.

Related

joining string from set reverses order

I have the string 'ABBA'. If I turn it into a set, I get this:
In: set('ABBA')
Out: {'A', 'B'}
However, if I join them as a string, it reverses the order:
In: ''.join(set('ABBA'))
Out: 'BA'
The same happens if I try to turn the string into a list:
In: list(set('ABBA'))
Out: ['B', 'A']
Why is this happening and how do I address it?
EDIT
The reason applying sorted doesn't work is that if I make a set out of 'CDA', it will return 'ACD', thus losing the order of the original string. My question pertains to preserving the original order of the set itself.
Sets are unordered collection i.e. you will get a different order every time you run the command and sets also have unique elements therefore there will be no repetition in the elements of the set.
if you try running this command for a few times set('ABBA') sometimes you will get the output as {'A', 'B'} and sometimes as {'B', 'A'} and that what happens when you are using the join command the output is sometimes taken as BA and sometimes it will show AB.
There is an ordered set recipe for this which is referred to from the Python 2 Documentation. This runs on Py2.6 or later and 3.0 or later without any modifications. The interface is almost exactly the same as a normal set, except that initialisation should be done with a list.
OrderedSet([1, 2, 3])
This is a MutableSet, so the signature for .union doesn't match that of set, but since it includes or something similar can easily be added
b = "AAEEBBCCDD"
a = set(b)#unordered
print(a)#{'B', 'D', 'C', 'A', 'E'}/{'A', 'E', 'B', 'D', 'C'}/,,,
#do not have reverses the order,only random
print(''.join(a))
print(list(a))
print(sorted(a, key=b.index))#Save original sequence(b)

I am getting SyntaxWarning: "is" with a literal. Did you mean "=="? [duplicate]

Two string variables are set to the same value. s1 == s2 always returns True, but s1 is s2 sometimes returns False.
If I open my Python interpreter and do the same is comparison, it succeeds:
>>> s1 = 'text'
>>> s2 = 'text'
>>> s1 is s2
True
Why is this?
is is identity testing, and == is equality testing. What happens in your code would be emulated in the interpreter like this:
>>> a = 'pub'
>>> b = ''.join(['p', 'u', 'b'])
>>> a == b
True
>>> a is b
False
So, no wonder they're not the same, right?
In other words: a is b is the equivalent of id(a) == id(b)
Other answers here are correct: is is used for identity comparison, while == is used for equality comparison. Since what you care about is equality (the two strings should contain the same characters), in this case the is operator is simply wrong and you should be using == instead.
The reason is works interactively is that (most) string literals are interned by default. From Wikipedia:
Interned strings speed up string
comparisons, which are sometimes a
performance bottleneck in applications
(such as compilers and dynamic
programming language runtimes) that
rely heavily on hash tables with
string keys. Without interning,
checking that two different strings
are equal involves examining every
character of both strings. This is
slow for several reasons: it is
inherently O(n) in the length of the
strings; it typically requires reads
from several regions of memory, which
take time; and the reads fills up the
processor cache, meaning there is less
cache available for other needs. With
interned strings, a simple object
identity test suffices after the
original intern operation; this is
typically implemented as a pointer
equality test, normally just a single
machine instruction with no memory
reference at all.
So, when you have two string literals (words that are literally typed into your program source code, surrounded by quotation marks) in your program that have the same value, the Python compiler will automatically intern the strings, making them both stored at the same memory location. (Note that this doesn't always happen, and the rules for when this happens are quite convoluted, so please don't rely on this behavior in production code!)
Since in your interactive session both strings are actually stored in the same memory location, they have the same identity, so the is operator works as expected. But if you construct a string by some other method (even if that string contains exactly the same characters), then the string may be equal, but it is not the same string -- that is, it has a different identity, because it is stored in a different place in memory.
The is keyword is a test for object identity while == is a value comparison.
If you use is, the result will be true if and only if the object is the same object. However, == will be true any time the values of the object are the same.
One last thing to note is you may use the sys.intern function to ensure that you're getting a reference to the same string:
>>> from sys import intern
>>> a = intern('a')
>>> a2 = intern('a')
>>> a is a2
True
As pointed out in previous answers, you should not be using is to determine equality of strings. But this may be helpful to know if you have some kind of weird requirement to use is.
Note that the intern function used to be a built-in on Python 2, but it was moved to the sys module in Python 3.
is is identity testing and == is equality testing. This means is is a way to check whether two things are the same things, or just equivalent.
Say you've got a simple person object. If it is named 'Jack' and is '23' years old, it's equivalent to another 23-year-old Jack, but it's not the same person.
class Person(object):
def __init__(self, name, age):
self.name = name
self.age = age
def __eq__(self, other):
return self.name == other.name and self.age == other.age
jack1 = Person('Jack', 23)
jack2 = Person('Jack', 23)
jack1 == jack2 # True
jack1 is jack2 # False
They're the same age, but they're not the same instance of person. A string might be equivalent to another, but it's not the same object.
This is a side note, but in idiomatic Python, you will often see things like:
if x is None:
# Some clauses
This is safe, because there is guaranteed to be one instance of the Null Object (i.e., None).
If you're not sure what you're doing, use the '=='.
If you have a little more knowledge about it you can use 'is' for known objects like 'None'.
Otherwise, you'll end up wondering why things doesn't work and why this happens:
>>> a = 1
>>> b = 1
>>> b is a
True
>>> a = 6000
>>> b = 6000
>>> b is a
False
I'm not even sure if some things are guaranteed to stay the same between different Python versions/implementations.
From my limited experience with Python, is is used to compare two objects to see if they are the same object as opposed to two different objects with the same value. == is used to determine if the values are identical.
Here is a good example:
>>> s1 = u'public'
>>> s2 = 'public'
>>> s1 is s2
False
>>> s1 == s2
True
s1 is a Unicode string, and s2 is a normal string. They are not the same type, but they are the same value.
I think it has to do with the fact that, when the 'is' comparison evaluates to false, two distinct objects are used. If it evaluates to true, that means internally it's using the same exact object and not creating a new one, possibly because you created them within a fraction of 2 or so seconds and because there isn't a large time gap in between it's optimized and uses the same object.
This is why you should be using the equality operator ==, not is, to compare the value of a string object.
>>> s = 'one'
>>> s2 = 'two'
>>> s is s2
False
>>> s2 = s2.replace('two', 'one')
>>> s2
'one'
>>> s2 is s
False
>>>
In this example, I made s2, which was a different string object previously equal to 'one' but it is not the same object as s, because the interpreter did not use the same object as I did not initially assign it to 'one', if I had it would have made them the same object.
The == operator tests value equivalence. The is operator tests object identity, and Python tests whether the two are really the same object (i.e., live at the same address in memory).
>>> a = 'banana'
>>> b = 'banana'
>>> a is b
True
In this example, Python only created one string object, and both a and b refers to it. The reason is that Python internally caches and reuses some strings as an optimization. There really is just a string 'banana' in memory, shared by a and b. To trigger the normal behavior, you need to use longer strings:
>>> a = 'a longer banana'
>>> b = 'a longer banana'
>>> a == b, a is b
(True, False)
When you create two lists, you get two objects:
>>> a = [1, 2, 3]
>>> b = [1, 2, 3]
>>> a is b
False
In this case we would say that the two lists are equivalent, because they have the same elements, but not identical, because they are not the same object. If two objects are identical, they are also equivalent, but if they are equivalent, they are not necessarily identical.
If a refers to an object and you assign b = a, then both variables refer to the same object:
>>> a = [1, 2, 3]
>>> b = a
>>> b is a
True
Reference: Think Python 2e by Allen B. Downey
I believe that this is known as "interned" strings. Python does this, so does Java, and so do C and C++ when compiling in optimized modes.
If you use two identical strings, instead of wasting memory by creating two string objects, all interned strings with the same contents point to the same memory.
This results in the Python "is" operator returning True because two strings with the same contents are pointing at the same string object. This will also happen in Java and in C.
This is only useful for memory savings though. You cannot rely on it to test for string equality, because the various interpreters and compilers and JIT engines cannot always do it.
Actually, the is operator checks for identity and == operator checks for equality.
From the language reference:
Types affect almost all aspects of object behavior. Even the importance of object identity is affected in some sense: for immutable types, operations that compute new values may actually return a reference to any existing object with the same type and value, while for mutable objects this is not allowed. E.g., after a = 1; b = 1, a and b may or may not refer to the same object with the value one, depending on the implementation, but after c = []; d = [], c and d are guaranteed to refer to two different, unique, newly created empty lists. (Note that c = d = [] assigns the same object to both c and d.)
So from the above statement we can infer that the strings, which are immutable types, may fail when checked with "is" and may succeed when checked with "is".
The same applies for int and tuple which are also immutable types.
is will compare the memory location. It is used for object-level comparison.
== will compare the variables in the program. It is used for checking at a value level.
is checks for address level equivalence
== checks for value level equivalence
is is identity testing and == is equality testing (see the Python documentation).
In most cases, if a is b, then a == b. But there are exceptions, for example:
>>> nan = float('nan')
>>> nan is nan
True
>>> nan == nan
False
So, you can only use is for identity tests, never equality tests.
The basic concept, we have to be clear, while approaching this question, is to understand the difference between is and ==.
"is" is will compare the memory location. if id(a)==id(b), then a is b returns true else it returns false.
So, we can say that is is used for comparing memory locations. Whereas,
== is used for equality testing which means that it just compares only the resultant values. The below shown code may acts as an example to the above given theory.
Code
In the case of string literals (strings without getting assigned to variables), the memory address will be same as shown in the picture. so, id(a)==id(b). The remaining of this is self-explanatory.

Cleanest way to declare a tuple of one string

Declaring a tuple of one string using the function produces one element per item in the string
(Pdb) tuple('VERSION',)
('V', 'E', 'R', 'S', 'I', 'O', 'N')
Declaring a tuple using commas feels like a side effect and I feel like is easy to miss.
(Pdb) ('VERSION',)
('VERSION',)
Is there a cleaner way to make a declaration like this?
For context I'm using a tuple of tuples and I'm iterating on all of the individual values. Rather than special case the single values I'm just making them a tuple of one item.
Edit: I see I was unclear about this.
I don't personally like the declaration of ('VERSION',) so I tried
(Pdb) tuple('VERSION',)
('V', 'E', 'R', 'S', 'I', 'O', 'N')`
And found that the function declaration to have this behavior.
I was interested to find that you could enclose the tuple declaration with a tuple declaration and that works.
(Pdb) tuple(['VERSION'])
('VERSION',)
(Pdb) tuple(('VERSION',))
('VERSION',)
Well, it's really a good question, how should we distinguish the difference between two parentheses and a tuple with one element, look at the following example :
>>> a = (1)+(2)
>>> a
3
>>> b = (1,)+(2,)
>>> b
(1, 2)
it's the beauty of the syntax of python, that , may look like an extra thing, but it makes difference () operator and tuples. so you should use it when you create a tuple with length one.
If you want to call the tuple by name, you can write tuple(['foo']) or tuple(('foo',)).
The tuple constructor takes an iterable, and, unless it's already a tuple, makes its elements the tuple's elements. There's no way around that.
The idiomatic way to define a 1-tuple is (foo,). I personally find the ,) a sign which is hard to miss, but tastes vary.
For ultimate clarity, just make your own function:
def tu(value):
return (value,)
Now you can tu('foo') and get a 1-tuple.

Can't compare strings with symbols? [duplicate]

Two string variables are set to the same value. s1 == s2 always returns True, but s1 is s2 sometimes returns False.
If I open my Python interpreter and do the same is comparison, it succeeds:
>>> s1 = 'text'
>>> s2 = 'text'
>>> s1 is s2
True
Why is this?
is is identity testing, and == is equality testing. What happens in your code would be emulated in the interpreter like this:
>>> a = 'pub'
>>> b = ''.join(['p', 'u', 'b'])
>>> a == b
True
>>> a is b
False
So, no wonder they're not the same, right?
In other words: a is b is the equivalent of id(a) == id(b)
Other answers here are correct: is is used for identity comparison, while == is used for equality comparison. Since what you care about is equality (the two strings should contain the same characters), in this case the is operator is simply wrong and you should be using == instead.
The reason is works interactively is that (most) string literals are interned by default. From Wikipedia:
Interned strings speed up string
comparisons, which are sometimes a
performance bottleneck in applications
(such as compilers and dynamic
programming language runtimes) that
rely heavily on hash tables with
string keys. Without interning,
checking that two different strings
are equal involves examining every
character of both strings. This is
slow for several reasons: it is
inherently O(n) in the length of the
strings; it typically requires reads
from several regions of memory, which
take time; and the reads fills up the
processor cache, meaning there is less
cache available for other needs. With
interned strings, a simple object
identity test suffices after the
original intern operation; this is
typically implemented as a pointer
equality test, normally just a single
machine instruction with no memory
reference at all.
So, when you have two string literals (words that are literally typed into your program source code, surrounded by quotation marks) in your program that have the same value, the Python compiler will automatically intern the strings, making them both stored at the same memory location. (Note that this doesn't always happen, and the rules for when this happens are quite convoluted, so please don't rely on this behavior in production code!)
Since in your interactive session both strings are actually stored in the same memory location, they have the same identity, so the is operator works as expected. But if you construct a string by some other method (even if that string contains exactly the same characters), then the string may be equal, but it is not the same string -- that is, it has a different identity, because it is stored in a different place in memory.
The is keyword is a test for object identity while == is a value comparison.
If you use is, the result will be true if and only if the object is the same object. However, == will be true any time the values of the object are the same.
One last thing to note is you may use the sys.intern function to ensure that you're getting a reference to the same string:
>>> from sys import intern
>>> a = intern('a')
>>> a2 = intern('a')
>>> a is a2
True
As pointed out in previous answers, you should not be using is to determine equality of strings. But this may be helpful to know if you have some kind of weird requirement to use is.
Note that the intern function used to be a built-in on Python 2, but it was moved to the sys module in Python 3.
is is identity testing and == is equality testing. This means is is a way to check whether two things are the same things, or just equivalent.
Say you've got a simple person object. If it is named 'Jack' and is '23' years old, it's equivalent to another 23-year-old Jack, but it's not the same person.
class Person(object):
def __init__(self, name, age):
self.name = name
self.age = age
def __eq__(self, other):
return self.name == other.name and self.age == other.age
jack1 = Person('Jack', 23)
jack2 = Person('Jack', 23)
jack1 == jack2 # True
jack1 is jack2 # False
They're the same age, but they're not the same instance of person. A string might be equivalent to another, but it's not the same object.
This is a side note, but in idiomatic Python, you will often see things like:
if x is None:
# Some clauses
This is safe, because there is guaranteed to be one instance of the Null Object (i.e., None).
If you're not sure what you're doing, use the '=='.
If you have a little more knowledge about it you can use 'is' for known objects like 'None'.
Otherwise, you'll end up wondering why things doesn't work and why this happens:
>>> a = 1
>>> b = 1
>>> b is a
True
>>> a = 6000
>>> b = 6000
>>> b is a
False
I'm not even sure if some things are guaranteed to stay the same between different Python versions/implementations.
From my limited experience with Python, is is used to compare two objects to see if they are the same object as opposed to two different objects with the same value. == is used to determine if the values are identical.
Here is a good example:
>>> s1 = u'public'
>>> s2 = 'public'
>>> s1 is s2
False
>>> s1 == s2
True
s1 is a Unicode string, and s2 is a normal string. They are not the same type, but they are the same value.
I think it has to do with the fact that, when the 'is' comparison evaluates to false, two distinct objects are used. If it evaluates to true, that means internally it's using the same exact object and not creating a new one, possibly because you created them within a fraction of 2 or so seconds and because there isn't a large time gap in between it's optimized and uses the same object.
This is why you should be using the equality operator ==, not is, to compare the value of a string object.
>>> s = 'one'
>>> s2 = 'two'
>>> s is s2
False
>>> s2 = s2.replace('two', 'one')
>>> s2
'one'
>>> s2 is s
False
>>>
In this example, I made s2, which was a different string object previously equal to 'one' but it is not the same object as s, because the interpreter did not use the same object as I did not initially assign it to 'one', if I had it would have made them the same object.
The == operator tests value equivalence. The is operator tests object identity, and Python tests whether the two are really the same object (i.e., live at the same address in memory).
>>> a = 'banana'
>>> b = 'banana'
>>> a is b
True
In this example, Python only created one string object, and both a and b refers to it. The reason is that Python internally caches and reuses some strings as an optimization. There really is just a string 'banana' in memory, shared by a and b. To trigger the normal behavior, you need to use longer strings:
>>> a = 'a longer banana'
>>> b = 'a longer banana'
>>> a == b, a is b
(True, False)
When you create two lists, you get two objects:
>>> a = [1, 2, 3]
>>> b = [1, 2, 3]
>>> a is b
False
In this case we would say that the two lists are equivalent, because they have the same elements, but not identical, because they are not the same object. If two objects are identical, they are also equivalent, but if they are equivalent, they are not necessarily identical.
If a refers to an object and you assign b = a, then both variables refer to the same object:
>>> a = [1, 2, 3]
>>> b = a
>>> b is a
True
Reference: Think Python 2e by Allen B. Downey
I believe that this is known as "interned" strings. Python does this, so does Java, and so do C and C++ when compiling in optimized modes.
If you use two identical strings, instead of wasting memory by creating two string objects, all interned strings with the same contents point to the same memory.
This results in the Python "is" operator returning True because two strings with the same contents are pointing at the same string object. This will also happen in Java and in C.
This is only useful for memory savings though. You cannot rely on it to test for string equality, because the various interpreters and compilers and JIT engines cannot always do it.
Actually, the is operator checks for identity and == operator checks for equality.
From the language reference:
Types affect almost all aspects of object behavior. Even the importance of object identity is affected in some sense: for immutable types, operations that compute new values may actually return a reference to any existing object with the same type and value, while for mutable objects this is not allowed. E.g., after a = 1; b = 1, a and b may or may not refer to the same object with the value one, depending on the implementation, but after c = []; d = [], c and d are guaranteed to refer to two different, unique, newly created empty lists. (Note that c = d = [] assigns the same object to both c and d.)
So from the above statement we can infer that the strings, which are immutable types, may fail when checked with "is" and may succeed when checked with "is".
The same applies for int and tuple which are also immutable types.
is will compare the memory location. It is used for object-level comparison.
== will compare the variables in the program. It is used for checking at a value level.
is checks for address level equivalence
== checks for value level equivalence
is is identity testing and == is equality testing (see the Python documentation).
In most cases, if a is b, then a == b. But there are exceptions, for example:
>>> nan = float('nan')
>>> nan is nan
True
>>> nan == nan
False
So, you can only use is for identity tests, never equality tests.
The basic concept, we have to be clear, while approaching this question, is to understand the difference between is and ==.
"is" is will compare the memory location. if id(a)==id(b), then a is b returns true else it returns false.
So, we can say that is is used for comparing memory locations. Whereas,
== is used for equality testing which means that it just compares only the resultant values. The below shown code may acts as an example to the above given theory.
Code
In the case of string literals (strings without getting assigned to variables), the memory address will be same as shown in the picture. so, id(a)==id(b). The remaining of this is self-explanatory.

How to return a value of a key from a dictionary in python

Let me have a dictionary:
P={'S':['dB','N'],'B':['C','CcB','bA']}
How can I get second value o the second key from dictionary P ?
Also, if the value is a string with more than one character like 'bA' (third value of key 'B'), can I somehow return first character of this value ?
Like #jonrsharpe has stated before, dictionaries aren't ordered by design.
What this means, is that everytime you attempt to access a dictionary "by order" you may encounter a different result.
Observe the following (python interactive interpreter):
>>>P={'S':['dB','N'],'B':['C','CcB','bA'], 'L':["qqq"]}
>>>P.keys()
['S', 'B', 'L']
Its easy to see that in this notice, the "order" as we defined is, matches the order that we receive from the Dictionary.keys() function.
However, you may also observe this result:
>>> P={'S':['dB','N'],'B':['C','CcB','bA'], 'L':["qqq"], 'A':[]}
>>> P.keys()
['A', 'S', 'B', 'L']
In this example, the value 'A' should be fourth in our list, but, it is actually the first value.
This is just a small example why you may not treat dictionaries as ordered lists.
Maybe you could go ahead and tell us what your intentions are and an alternative may be suggested.

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