How to compare two lists in python

Question

Suppose I have two lists (or numpy.arrays):

a = [1,2,3]
b = [4,5,6]

How can I check if each element of a is smaller than corresponding element of b at the same index? (I am assuming indices are starting from 0) i.e.

at index 0 value of a = 1 < value of b = 4
at index 1 value of a = 2 < value of b = 5
at index 2 value of a = 3 < value of b = 6

If a were equal to [1,2,7], then that would be incorrect because at index 2 value of a is greater than that of b. Also if a's length were any smaller than that of b, it should be comparing only the indices of a with those of b.

For example this pair a, b

a = [1,2]
b = [3,4,5]

at indices 0 and 1, value of a is smaller than b, thus this would also pass the check.

P.S.--> I have to use the above conditions inside a if statement. And also, no element of a should be equal to that of b i.e. strictly lesser. Feel free to use as many as tools as you like. (Although I am using lists here, you can convert the above lists into numpy arrays too.)

Answer 1

Answering both parts with zip and all

all(i < j for (i, j) in zip(a, b))

zip will pair the values from the beginning of a with values from beginning of b; the iteration ends when the shorter iterable has run out. all returns True if and only if all items in a given are true in boolean context. Also, when any item fails, False will be returned early.

Example results:

>>> a = [1,2,3]
>>> b = [4,5,6]
>>> all(i < j for (i, j) in zip(a, b))
True
>>> a = [1,2,7]
>>> b = [4,5,6]
>>> all(i < j for (i, j) in zip(a, b))
False
>>> a = [1,2]
>>> b = [4,5,-10]
>>> all(i < j for (i, j) in zip(a, b))
True

Timings with IPython 3.4.2:

In [1]: a = [1] * 10000
In [2]: b = [1] * 10000
In [3]: %timeit all(i < j for (i, j) in zip(a, b))
1000 loops, best of 3: 995 µs per loop
In [4]: %timeit all(starmap(lt, zip(a, b)))
1000 loops, best of 3: 487 µs per loop

So the starmap is faster in this case. In general 2 things are relatively slow in Python: function calls and global name lookups. The starmap of Retard's solution seems to win here exactly because the tuple yielded from zip can be fed as-is as the *args to the lt builtin function, whereas my code needs to deconstruct it.