What does eval(dir()[0]) do in python

Question

I came across a solution for a program in python but I couldn't understand what it does even after searching. Could someone please explain me what this statement will do.

a, b, c = map(numpy.array,eval(dir()[0]))

Answer 1

Function dir, when called without arguments returns the names of all local variables, similar to locals().keys().

def f(y):
     print(dir())  # prints ['y']

Then, obviously, dir()[0] is the name of the first of the local variables and eval(dir()[0]) evaluates the variable name, i.e. returns the first local variable's value.

def f(y):
     print(dir())  # prints ['y']
     print(dir()[0])  # prints 'y'
     print(eval(dir()[0]))  # prints the value of y

For example:

>>> f(77)
['y']
y
77
>>> f([1,2,3])
['y']
y
[1, 2, 3]

Function map calls the first argument (which has to be callable) with each of the values in the second argument (which has to be iterable), and generates the results e.g.

>>> for result in map(str.upper, ['foo', 'bar', 'baz']):
...     print(result)
...
FOO
BAR
BAZ

Combining those together, and assuming that the first local variable is a list named first_variable, then this code:

a, b, c = map(numpy.array,eval(dir()[0]))

would be the same as this code:

a, b, c = first_variable
a = numpy.array(a)
b = numpy.array(b)
c = numpy.array(c)