'dict_items' object does not support indexing [duplicate] - python

def shuffle(self, x, random=None, int=int):
"""x, random=random.random -> shuffle list x in place; return None.
Optional arg random is a 0-argument function returning a random
float in [0.0, 1.0); by default, the standard random.random.
"""
randbelow = self._randbelow
for i in reversed(range(1, len(x))):
# pick an element in x[:i+1] with which to exchange x[i]
j = randbelow(i+1) if random is None else int(random() * (i+1))
x[i], x[j] = x[j], x[i]
When I run the shuffle function it raises the following error, why is that?
TypeError: 'dict_keys' object does not support indexing

Clearly you're passing in d.keys() to your shuffle function. Probably this was written with python2.x (when d.keys() returned a list). With python3.x, d.keys() returns a dict_keys object which behaves a lot more like a set than a list. As such, it can't be indexed.
The solution is to pass list(d.keys()) (or simply list(d)) to shuffle.

You're passing the result of somedict.keys() to the function. In Python 3, dict.keys doesn't return a list, but a set-like object that represents a view of the dictionary's keys and (being set-like) doesn't support indexing.
To fix the problem, use list(somedict.keys()) to collect the keys, and work with that.

Convert an iterable to a list may have a cost. Instead, to get the the first item, you can use:
next(iter(keys))
Or, if you want to iterate over all items, you can use:
items = iter(keys)
while True:
try:
item = next(items)
except StopIteration as e:
pass # finish

In Python 2 dict.keys() return a list, whereas in Python 3 it returns a generator.
You could only iterate over it's values else you may have to explicitly convert it to a list i.e. pass it to a list function.

Why you need to implement shuffle when it already exists? Stay on the shoulders of giants.
import random
d1 = {0:'zero', 1:'one', 2:'two', 3:'three', 4:'four',
5:'five', 6:'six', 7:'seven', 8:'eight', 9:'nine'}
keys = list(d1)
random.shuffle(keys)
d2 = {}
for key in keys: d2[key] = d1[key]
print(d1)
print(d2)

Related

Python - how to take the max length from a value of dictionary? Without lambda

Let us say I have this function:
def frequent_base(self):
dict = {}
for i in range(len(self.items)):
if self.items[i].base not in dict:
dict[self.items[i].base] = [(self.items[i].value)]
else:
dict[self.items[i].base] += [(self.items[i].base)]
return max(len(dict[self.items]), key=len(d))
Now, I can make it complicated and build a function which returns me index and such..
but it is bad coding and bad habit and takes long time ( especially in a test ).
How do I take the length?
let us say I have:
key1 with length 3 of value ( key1 has 3 values )
key2 with length 4 of value ( key 2 has 4 values )
key3 with length 2 of value ( key 3 has 2 values )
How do I take, not the key itself, not the value itself, but the len of values of key? which is 4 in this case.
or how do I take the key itself and then say length of value of that key? But I want to use Max function, I need to understand how to use that function good, with the key.
I will write and make myself super clear:
dict[1] = [1,2,3]
dict[2] = [1,2,3,4,5]
dict[3] = [1,2,3,7,8,9,10]
dict = {1: [1,2,3], 2:[1,2,3,4,5], 3:[1,2,3,7,8,9,10]}
I wish to return not dict[3], not 3, not the list of dict[3] it self.
I wish to return the length of the dict[3], which is 7
def frequent_base(self):
dict = {}
for i in range(len(self.items)):
if self.items[i].base not in dict:
dict[self.items[i].base] = [(self.items[i].value)]
else:
dict[self.items[i].base] += [(self.items[i].base)]
def key_for_len(dictionary):
return dictionary[1]
return max(dict.items(), key= key_for_len)
I am received error
Only thing you seem to need is maximal length amongst the values of your dictionary. You can easily get all the values using d.values() (d.items() would give you (key, value) tuples, which are harder to compare). Now we can easily calculate lengths of each value with generator comprehension (very much like list comprehension, which would also work) - len(v) for v in d.values(). Now we have an iterable with all the lengths, so it's just a matter of finding the maximum -
max(len(v) for v in d.values())
Should you need to get key or value with maximum length, we'd have to take a slightly different approach - we can use max key = argument to specify how we decide which element in the iterable is maximal - it is obvious when we are trying to get a maximum from few numbers, but not when we try to decide if (1, 3) is bigger than (2, 2) - in such case, we need to create a function that maps our items to easily comparable things like a number. In this case, we'd have tuples of (key, value) and we are comparing length of value - thus our function would be
def lenOfValue(kv):
return len(kv[1]) # kv[1] - 2nd element of a (key, value) tuple
code(1)
Then we pass that to max:
print(max(d.items(), key = lenOfValue))
And we get (2, [3, 4, 5, 6])
Bonus: lambda
Lambdas can be used here, which are really just a shorthand that lets us skip defining whole another function that we will probably never use again.
This code would be functionally exactly the same.
print(max(d.items(), key = lambda kv: len(kv[1])))
code(2)
Lambdas are nothing very complicated, being just a notation for creating simple, one-liner functions without all the bother of a def block, return etc.
Because Python's functions are objects like nearly anything else, this piece of code:
lenOfValue = lambda kv: len(kv[1])
really is in no way different that our previously used more lengthy definition of:
def lenOfValue(kv):
return len(kv[1])
It saves us few words and shows us the middle step between code(1) and code(2).
a_dict = {'some_key':[67,30,10], 'another_key':[87]}
max({ (k,len(v)) for k,v in a_dict.items() })
('some_key', 3)

How to compare two list of dictionaries [duplicate]

a = [1, 2, 3, 1, 2, 3]
b = [3, 2, 1, 3, 2, 1]
a & b should be considered equal, because they have exactly the same elements, only in different order.
The thing is, my actual lists will consist of objects (my class instances), not integers.
O(n): The Counter() method is best (if your objects are hashable):
def compare(s, t):
return Counter(s) == Counter(t)
O(n log n): The sorted() method is next best (if your objects are orderable):
def compare(s, t):
return sorted(s) == sorted(t)
O(n * n): If the objects are neither hashable, nor orderable, you can use equality:
def compare(s, t):
t = list(t) # make a mutable copy
try:
for elem in s:
t.remove(elem)
except ValueError:
return False
return not t
You can sort both:
sorted(a) == sorted(b)
A counting sort could also be more efficient (but it requires the object to be hashable).
>>> from collections import Counter
>>> a = [1, 2, 3, 1, 2, 3]
>>> b = [3, 2, 1, 3, 2, 1]
>>> print (Counter(a) == Counter(b))
True
If you know the items are always hashable, you can use a Counter() which is O(n)
If you know the items are always sortable, you can use sorted() which is O(n log n)
In the general case you can't rely on being able to sort, or has the elements, so you need a fallback like this, which is unfortunately O(n^2)
len(a)==len(b) and all(a.count(i)==b.count(i) for i in a)
If you have to do this in tests:
https://docs.python.org/3.5/library/unittest.html#unittest.TestCase.assertCountEqual
assertCountEqual(first, second, msg=None)
Test that sequence first contains the same elements as second, regardless of their order. When they don’t, an error message listing the differences between the sequences will be generated.
Duplicate elements are not ignored when comparing first and second. It verifies whether each element has the same count in both sequences. Equivalent to: assertEqual(Counter(list(first)), Counter(list(second))) but works with sequences of unhashable objects as well.
New in version 3.2.
or in 2.7:
https://docs.python.org/2.7/library/unittest.html#unittest.TestCase.assertItemsEqual
Outside of tests I would recommend the Counter method.
The best way to do this is by sorting the lists and comparing them. (Using Counter won't work with objects that aren't hashable.) This is straightforward for integers:
sorted(a) == sorted(b)
It gets a little trickier with arbitrary objects. If you care about object identity, i.e., whether the same objects are in both lists, you can use the id() function as the sort key.
sorted(a, key=id) == sorted(b, key==id)
(In Python 2.x you don't actually need the key= parameter, because you can compare any object to any object. The ordering is arbitrary but stable, so it works fine for this purpose; it doesn't matter what order the objects are in, only that the ordering is the same for both lists. In Python 3, though, comparing objects of different types is disallowed in many circumstances -- for example, you can't compare strings to integers -- so if you will have objects of various types, best to explicitly use the object's ID.)
If you want to compare the objects in the list by value, on the other hand, first you need to define what "value" means for the objects. Then you will need some way to provide that as a key (and for Python 3, as a consistent type). One potential way that would work for a lot of arbitrary objects is to sort by their repr(). Of course, this could waste a lot of extra time and memory building repr() strings for large lists and so on.
sorted(a, key=repr) == sorted(b, key==repr)
If the objects are all your own types, you can define __lt__() on them so that the object knows how to compare itself to others. Then you can just sort them and not worry about the key= parameter. Of course you could also define __hash__() and use Counter, which will be faster.
If the comparison is to be performed in a testing context, use assertCountEqual(a, b) (py>=3.2) and assertItemsEqual(a, b) (2.7<=py<3.2).
Works on sequences of unhashable objects too.
If the list contains items that are not hashable (such as a list of objects) you might be able to use the Counter Class and the id() function such as:
from collections import Counter
...
if Counter(map(id,a)) == Counter(map(id,b)):
print("Lists a and b contain the same objects")
Let a,b lists
def ass_equal(a,b):
try:
map(lambda x: a.pop(a.index(x)), b) # try to remove all the elements of b from a, on fail, throw exception
if len(a) == 0: # if a is empty, means that b has removed them all
return True
except:
return False # b failed to remove some items from a
No need to make them hashable or sort them.
I hope the below piece of code might work in your case :-
if ((len(a) == len(b)) and
(all(i in a for i in b))):
print 'True'
else:
print 'False'
This will ensure that all the elements in both the lists a & b are same, regardless of whether they are in same order or not.
For better understanding, refer to my answer in this question
You can write your own function to compare the lists.
Let's get two lists.
list_1=['John', 'Doe']
list_2=['Doe','Joe']
Firstly, we define an empty dictionary, count the list items and write in the dictionary.
def count_list(list_items):
empty_dict={}
for list_item in list_items:
list_item=list_item.strip()
if list_item not in empty_dict:
empty_dict[list_item]=1
else:
empty_dict[list_item]+=1
return empty_dict
After that, we'll compare both lists by using the following function.
def compare_list(list_1, list_2):
if count_list(list_1)==count_list(list_2):
return True
return False
compare_list(list_1,list_2)
from collections import defaultdict
def _list_eq(a: list, b: list) -> bool:
if len(a) != len(b):
return False
b_set = set(b)
a_map = defaultdict(lambda: 0)
b_map = defaultdict(lambda: 0)
for item1, item2 in zip(a, b):
if item1 not in b_set:
return False
a_map[item1] += 1
b_map[item2] += 1
return a_map == b_map
Sorting can be quite slow if the data is highly unordered (timsort is extra good when the items have some degree of ordering). Sorting both also requires fully iterating through both lists.
Rather than mutating a list, just allocate a set and do a left-->right membership check, keeping a count of how many of each item exist along the way:
If the two lists are not the same length you can short circuit and return False immediately.
If you hit any item in list a that isn't in list b you can return False
If you get through all items then you can compare the values of a_map and b_map to find out if they match.
This allows you to short-circuit in many cases long before you've iterated both lists.
plug in this:
def lists_equal(l1: list, l2: list) -> bool:
"""
import collections
compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
ref:
- https://stackoverflow.com/questions/9623114/check-if-two-unordered-lists-are-equal
- https://stackoverflow.com/questions/7828867/how-to-efficiently-compare-two-unordered-lists-not-sets
"""
compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
set_comp = set(l1) == set(l2) # removes duplicates, so returns true when not sometimes :(
multiset_comp = compare(l1, l2) # approximates multiset
return set_comp and multiset_comp #set_comp is gere in case the compare function doesn't work

comparing contents of two lists python [duplicate]

a = [1, 2, 3, 1, 2, 3]
b = [3, 2, 1, 3, 2, 1]
a & b should be considered equal, because they have exactly the same elements, only in different order.
The thing is, my actual lists will consist of objects (my class instances), not integers.
O(n): The Counter() method is best (if your objects are hashable):
def compare(s, t):
return Counter(s) == Counter(t)
O(n log n): The sorted() method is next best (if your objects are orderable):
def compare(s, t):
return sorted(s) == sorted(t)
O(n * n): If the objects are neither hashable, nor orderable, you can use equality:
def compare(s, t):
t = list(t) # make a mutable copy
try:
for elem in s:
t.remove(elem)
except ValueError:
return False
return not t
You can sort both:
sorted(a) == sorted(b)
A counting sort could also be more efficient (but it requires the object to be hashable).
>>> from collections import Counter
>>> a = [1, 2, 3, 1, 2, 3]
>>> b = [3, 2, 1, 3, 2, 1]
>>> print (Counter(a) == Counter(b))
True
If you know the items are always hashable, you can use a Counter() which is O(n)
If you know the items are always sortable, you can use sorted() which is O(n log n)
In the general case you can't rely on being able to sort, or has the elements, so you need a fallback like this, which is unfortunately O(n^2)
len(a)==len(b) and all(a.count(i)==b.count(i) for i in a)
If you have to do this in tests:
https://docs.python.org/3.5/library/unittest.html#unittest.TestCase.assertCountEqual
assertCountEqual(first, second, msg=None)
Test that sequence first contains the same elements as second, regardless of their order. When they don’t, an error message listing the differences between the sequences will be generated.
Duplicate elements are not ignored when comparing first and second. It verifies whether each element has the same count in both sequences. Equivalent to: assertEqual(Counter(list(first)), Counter(list(second))) but works with sequences of unhashable objects as well.
New in version 3.2.
or in 2.7:
https://docs.python.org/2.7/library/unittest.html#unittest.TestCase.assertItemsEqual
Outside of tests I would recommend the Counter method.
The best way to do this is by sorting the lists and comparing them. (Using Counter won't work with objects that aren't hashable.) This is straightforward for integers:
sorted(a) == sorted(b)
It gets a little trickier with arbitrary objects. If you care about object identity, i.e., whether the same objects are in both lists, you can use the id() function as the sort key.
sorted(a, key=id) == sorted(b, key==id)
(In Python 2.x you don't actually need the key= parameter, because you can compare any object to any object. The ordering is arbitrary but stable, so it works fine for this purpose; it doesn't matter what order the objects are in, only that the ordering is the same for both lists. In Python 3, though, comparing objects of different types is disallowed in many circumstances -- for example, you can't compare strings to integers -- so if you will have objects of various types, best to explicitly use the object's ID.)
If you want to compare the objects in the list by value, on the other hand, first you need to define what "value" means for the objects. Then you will need some way to provide that as a key (and for Python 3, as a consistent type). One potential way that would work for a lot of arbitrary objects is to sort by their repr(). Of course, this could waste a lot of extra time and memory building repr() strings for large lists and so on.
sorted(a, key=repr) == sorted(b, key==repr)
If the objects are all your own types, you can define __lt__() on them so that the object knows how to compare itself to others. Then you can just sort them and not worry about the key= parameter. Of course you could also define __hash__() and use Counter, which will be faster.
If the comparison is to be performed in a testing context, use assertCountEqual(a, b) (py>=3.2) and assertItemsEqual(a, b) (2.7<=py<3.2).
Works on sequences of unhashable objects too.
If the list contains items that are not hashable (such as a list of objects) you might be able to use the Counter Class and the id() function such as:
from collections import Counter
...
if Counter(map(id,a)) == Counter(map(id,b)):
print("Lists a and b contain the same objects")
Let a,b lists
def ass_equal(a,b):
try:
map(lambda x: a.pop(a.index(x)), b) # try to remove all the elements of b from a, on fail, throw exception
if len(a) == 0: # if a is empty, means that b has removed them all
return True
except:
return False # b failed to remove some items from a
No need to make them hashable or sort them.
I hope the below piece of code might work in your case :-
if ((len(a) == len(b)) and
(all(i in a for i in b))):
print 'True'
else:
print 'False'
This will ensure that all the elements in both the lists a & b are same, regardless of whether they are in same order or not.
For better understanding, refer to my answer in this question
You can write your own function to compare the lists.
Let's get two lists.
list_1=['John', 'Doe']
list_2=['Doe','Joe']
Firstly, we define an empty dictionary, count the list items and write in the dictionary.
def count_list(list_items):
empty_dict={}
for list_item in list_items:
list_item=list_item.strip()
if list_item not in empty_dict:
empty_dict[list_item]=1
else:
empty_dict[list_item]+=1
return empty_dict
After that, we'll compare both lists by using the following function.
def compare_list(list_1, list_2):
if count_list(list_1)==count_list(list_2):
return True
return False
compare_list(list_1,list_2)
from collections import defaultdict
def _list_eq(a: list, b: list) -> bool:
if len(a) != len(b):
return False
b_set = set(b)
a_map = defaultdict(lambda: 0)
b_map = defaultdict(lambda: 0)
for item1, item2 in zip(a, b):
if item1 not in b_set:
return False
a_map[item1] += 1
b_map[item2] += 1
return a_map == b_map
Sorting can be quite slow if the data is highly unordered (timsort is extra good when the items have some degree of ordering). Sorting both also requires fully iterating through both lists.
Rather than mutating a list, just allocate a set and do a left-->right membership check, keeping a count of how many of each item exist along the way:
If the two lists are not the same length you can short circuit and return False immediately.
If you hit any item in list a that isn't in list b you can return False
If you get through all items then you can compare the values of a_map and b_map to find out if they match.
This allows you to short-circuit in many cases long before you've iterated both lists.
plug in this:
def lists_equal(l1: list, l2: list) -> bool:
"""
import collections
compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
ref:
- https://stackoverflow.com/questions/9623114/check-if-two-unordered-lists-are-equal
- https://stackoverflow.com/questions/7828867/how-to-efficiently-compare-two-unordered-lists-not-sets
"""
compare = lambda x, y: collections.Counter(x) == collections.Counter(y)
set_comp = set(l1) == set(l2) # removes duplicates, so returns true when not sometimes :(
multiset_comp = compare(l1, l2) # approximates multiset
return set_comp and multiset_comp #set_comp is gere in case the compare function doesn't work

How do I properly pass an argument to a function

I'm from a C++ background so this problem seems a little absurd to me:
Let's suppose I have a function:
def scale(data, factor):
for val in data:
val *= factor
This doesn't work as intended, if I pass a list, it changes nothing, but
def scale(data, factor):
for index, val in enumerate(data):
data[index] *= factor
and lst = [val * factor for val in lst] works properly.
How does Python handle argument passing? How do I know if the actual reference, or alias is passed?
if you want to mutate the list, you need to reference the elements. This version uses map (it could be written using list comprehensions)
def scale(data, factor):
return map(lambda x : x*factor, data)
a lambda function is an anonymous function.
>>> (lambda x : x + 1) (5)
6
The x takes the place of the variable in this case 5 + 1
So in this case, we traverse the list applying the function f(x) -> x * factor to every element of the list. The original list is not mutated, but instead we return a new version.
In python basic data types are passed by value - for example int, str, bool etc are passed by value
Derived data types like classes, enum, list, dict are passed by reference.
In your example, the problem is how you use the for loop - not the function argument. If you do:
for val in lst:
val += 1
The values inside lst won't get updated because the val is not the same as lst[0], lst[1] and so on IF val is of the basic data types. So, even here, the val is copied by value.
Second, In your example with enumerate:
But when you loop over the enumerated list, you are using data[index] - which modifies the element in the actual list.
And finally, In your example with the generator:
lst = [val * factor for val in lst] - here the generator loops over every element and creates a new list which is again stored in lst. This is something like a = a + 2 but extended to lists.
This behaviour is so because the basic data types are passed by value and the derived data types like lists are passed by reference consider this
>>> x = 24
>>> x + 1
25
>>> x
24
but on the otherhand with a list
>>> y = [1, 2, 3, 4]
>>> y.remove(2)
>>> y
[1,3,4]
so you should always be careful to reassign values back when performing operations on them in the case of the basic data ypes and also be careful with datatypes that are passed by reference because you could accidentally modify a variable without knowing

Sum a list which contains 'None' using Python

Basically my question is say you have an list containing 'None' how would you try retrieving the sum of the list. Below is an example I tried which doesn't work and I get the error: TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'. Thanks
def sumImport(self):
my_list = [[1,2,3,None],[1,2,3],[1,1],[1,1,2,2]]
k = sum(chain.from_iterable(my_list))
return k
You can use filter function
>>> sum(filter(None, [1,2,3,None]))
6
Updated from comments
Typically filter usage is filter(func, iterable), but passing None as first argument is a special case, described in Python docs. Quoting:
If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.
Remove None (and zero) elements before summing by using filter:
>>> k = sum(filter(None, chain.from_iterable(my_list)))
>>> k
20
To see why this works, see the documentation for filter:
filter(function, iterable)
Construct a list from those elements of iterable for which function returns true. iterable may be either a sequence, a container which supports iteration, or an iterator. If iterable is a string or a tuple, the result also has that type; otherwise it is always a list. If function is None, the identity function is assumed, that is, all elements of iterable that are false are removed.
Note that filter(function, iterable) is equivalent to [item for item in iterable if function(item)] if function is not None and [item for item in iterable if item] if function is None.
Another suggestion:
from itertools import chain
k = sum(x for x in chain.from_iterable(my_list) if x)
Assuming you want to treat None as zero, a simple way is
sum(x if x is not None else 0 for x in chain.from_iterable(my_list))
Explicitly, this is equivalent to filter:
k = sum([x for x in chain.from_iterable(my_list) if x])
That saves me from remembering another function. :P
You always have the option of just writing the loop you want:
k = 0
for sublist in my_list:
for val in sublist:
if val is not None:
k += val
But it certainly doesn’t hurt to know about filter either.
Just using sum and map:
sum(map(lambda x: x or 0, [1,2,3,None]))
# 6

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