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How to do logical functions on arrays in python

Is it possible somehow to preform logical functions on arrays like eg.
a= [1,2,3,4,5,6]
b= [1,3,5,7]
c= a and b
resulting in c=[1,3,5]
so only the values that are present in both the arrays.
The same with or eg:
d = a OR b
resulting in b=[1,2,3,4,5,6,7]
Is this possible somehow in python or are there short functions for this?
Thanks for your response

Lists do not support logical operations , but you can do it in sets:
a= [1,2,3,4,5,6]
b= [1,3,5,7]
c = list(set(a) | set(b))
d = list(set(a) & set(b))
print(c)
# OUTPUT: [1, 2, 3, 4, 5, 6, 7]
print(d)
# OUTPUT: [1, 3, 5]

You probably should use python set if your collections contain unique values. They are actually mathematical sets: https://en.wikipedia.org/wiki/Set_(mathematics) and support general operation for sets such as intersection and union
>>> a = {1, 2, 3, 4, 5, 6}
>>> b = {1, 3, 5, 7}
>>> c = a.intersection(b)
>>> c
{1, 3, 5}

Here is your snippet:
def and_array(a, b):
return [a[i] for i in range(min(len(a), len(b))) if a[i] in b or b[i] in a]
def or_array(a, b):
return a + [element for element in b if not element in a]
a= [1,2,3,4,5,6]
b= [1,3,5,7]
c = and_array(a, b)
d = or_array(a, b)
print(c)
print(d)
Result:
[1, 2, 3]
[1, 2, 3, 4, 5, 6, 7]
It is not that fast, avoid this for very large lists and use numpy or build-in functions!

Take elements from multiple lists

Given multiple lists like the ones shown:
a = [1, 2, 3]
b = [5, 6, 7, 8]
c = [9, 0, 1]
d = [2, 3, 4, 5, 6, 7]
...
I want to be able to combine them to take as many elements from the first list as I can before starting to take elements from the second list, so the result would be:
result = [1, 2, 3, 8, 6, 7]
Is there a particularly nice way to write this? I can't think of a really simple one without a for loop. Maybe a list comprehension with a clever zip.

Simple slicing and concatenation:
a + b[len(a):]
Or with more lists:
res = []
for lst in (a, b, c, d):
res += lst[len(res):]
# [1, 2, 3, 8, 6, 7]

With itertools.zip_longest() for Python 3, works on any number of input lists:
>>> from itertools import zip_longest
>>> [next(x for x in t if x is not None) for t in zip_longest(a,b,c,d)]
[1, 2, 3, 8, 6, 7]
The default fill value is None so take the first none None element in each tuple created with the zip_longest call (you can change the defaults and criteria if None is a valid data value)

With functools.reduce:
from functools import reduce
print(list(reduce(lambda a, b: a + b[len(a):], [a, b, c, d])))
This outputs:
[1, 2, 3, 8, 6, 7]

How do you subtract list a from list b in Python?

So I have 2 lists, say list a and list b where
a = [9, 8, 7, 6]
b = [1, 2, 3, 4]
How would I go about subtracting the contents of b from a?

You can use the map function and it's feature to support more than one iterable (the following assumes Python2):
>>> a = [9, 8, 7, 6]
>>> b = [1, 2, 3, 4]
>>> map(lambda x,y: x-y, a, b)
[8, 6, 4, 2]
map applies the first argument (which has to be a function) on all elements of the following arguments. For example:
>>> from math import sqrt
>>> map(sqrt, [1,2,9])
[1.0, 1.4142135623730951, 3.0]
If you use more than two arguments, the function in the first parameter must take more parameters, because it is called with elements from each list:
>>> from math import pow
>>> map(pow, [2,3,4], [2,3,4])
[4.0, 27.0, 256.0]
The result is 2^2, 3^3 and 4^4.
The lambda in my example is just a shorter way to define the subtraction function, the following code would do the same:
def sub(x,y):
return x-y
map(sub, a, b)

You can zip the two lists and subtract the subelements to create a new list:
zip(b,a) -> [(1, 9), (2, 8), (3, 7), (4, 6)]
a = [9, 8, 7, 6]
b = [1, 2, 3, 4]
print([y-x for x,y in zip(b,a)])
[8, 6, 4, 2]
If you want to change a itself use enumerate subtracting elements at common indexes:
for ind,ele in enumerate(a):
a[ind] -= b[ind]
print(a)
[8, 6, 4, 2]
Or using numpy:
import numpy as np
a = np.array([9, 8, 7, 6])
b = np.array([1, 2, 3, 4])
print(a - b)
[8 6 4 2]

a=[1,2,3,4,5]
b=[9,8,7,6,4]
t=0
h=[]
lenA=len(a)
while lenA != t:
x=a[t]-b[t]
t=t+1
h.append(x)

How to create dynamic link between list variables

The dictionary's methods .viewvalues() and .viewkeys() allow to create the list variables that will be linked and dynamically updated on every dictionary modification such as:
diction={'one':1,'two':2,'three':3}
dict_values=dictVar.viewvalues()
dict_keys=dictVar.viewkeys()
I wonder if a similar functionality could be achieved with lists. So if there are two "source" list variables and a third list is a result of sums of twos:
a=[1,2,3]
b=[4,5,6]
sum=a+b
Now what i want is a list variable sum to get updated if/when list variable a or list variable b is modified. How to achieve that?

I'd define a function to do it and then call that whenever you need the list.
a=[1,2,3]
b=[4,5,6]
def sum(a, b):
return a + b
Then, in an interpreter:
>>> sum(a, b)
[1, 2, 3, 4, 5, 6]
>>> a.append(5)
>>> sum(a, b)
[1, 2, 3, 5, 4, 5, 6]
If it's not necessary that it be a flat list, you can easily do what you'd want.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> sum = [a, b]
>>> print(sum)
[[1, 2, 3], [4, 5, 6]]
>>> a.append(8)
>>> print(sum)
[[1, 2, 3, 8], [4, 5, 6]]
That said, I'd recommend against defining a variable named sum as it's a built-in Python function.

You could do it the other way around, using numpy arrays.
>>> import numpy as np
>>> ab = np.array([1,2,3,4,5,6])
>>> a = ab[:3]
>>> b = ab[3:]
>>> a, b
(array([1, 2, 3]), array([4, 5, 6]))
>>> a[1] = 9
>>> ab
array([1, 9, 3, 4, 5, 6])
>>> ab[0] = 7
>>> a
array([7, 9, 3])
Here, a and b are "views" on the array ab, and modifying one will also modify the other.
Starting with a and b, just create a numpy array from a+b and redefine a and b accordingly:
>>> a, b = [1,2,3], [4,5,6]
>>> ab = np.array(a+b)
>>> a, b = ab[:3], ab[3:]

You will have to right a custom data structure to do this. Here is something in the right direction...
class LinkedArrays(object):
def __init__(self, sourceArray1, sourceArray2, combineFunction):
self.sa1, self.sa2 = sourceArray1, sourceArray2
self.__combineFunction = combineFunction
self.__update()
def updateSourceArray1(self, index, value):
self.sa1[index] = value
self.__update()
def updateSourceArray2(self, index, value):
self.sa2[index] = value
self.__update()
def __update(self):
self.combinedArray = [self.__combineFunction(self.sa1[i], self.sa2[i]) for i in range(len(self.sa1))]
def __getitem__(self, item):
return self.combinedArray[item]
summedArrays = LinkedArrays([1, 2, 3], [4, 5, 6], lambda x, y: x+y)
print summedArrays[0] # print 5
summedArrays.updateSourceArray1(0, 6)
print summedArrays[0] # print 10

compare two lists in python and return indices of matched values

For two lists a and b, how can I get the indices of values that appear in both? For example,
a = [1, 2, 3, 4, 5]
b = [9, 7, 6, 5, 1, 0]
return_indices_of_a(a, b)
would return [0,4], with (a[0],a[4]) = (1,5).

The best way to do this would be to make b a set since you are only checking for membership inside it.
>>> a = [1, 2, 3, 4, 5]
>>> b = set([9, 7, 6, 5, 1, 0])
>>> [i for i, item in enumerate(a) if item in b]
[0, 4]

def return_indices_of_a(a, b):
b_set = set(b)
return [i for i, v in enumerate(a) if v in b_set]

For larger lists this may be of help:
for item in a:
index.append(bisect.bisect(b,item))
idx = np.unique(index).tolist()
Be sure to import numpy.

you can do it like this,with no need to set :
a = [1, 2, 3, 4, 5]
b = {9, 7, 6, 5, 1, 0}
[i for i, item in enumerate(a) if item in b]
inspired by #jamylak