Python2.x:

for idx in range(0, int(100 / 0.5)):

    print 0.5 * idx      

outputs:

0.0

0.5

1.0

1.5

..

99.0

99.5

Numpy:

numpy.arange would also do the trick.

numpy.arange(0, 100, 0.5)

The range objects are special:

Python will compare range objects as Sequences. What that essentially means is that the comparison doesn't evaluate how they represent a given sequence but rather what they represent.

The fact that the start, stop and step parameters are completely different plays no difference here because they all represent an empty list when expanded:

For example, the first range object:

list(range(0))  # []

and the second range object:

list(range(2, 2, 2)) # []

Both represent an empty list and since two empty lists compare equal (True) so will the range objects that represent them.

As a result, you can have completely different looking range objects; if they represent the same sequence they will compare equal:

range(1, 5, 100) == range(1, 30, 100) 

Both represent a list with a single element [1] so these two will also compare equal.

No, range objects are really special:

Do note, though, that even though the comparison doesn't evaluate how they represent a sequence the result of comparing can be achieved using solely the values of start, step along with the len of the range objects; this has very interesting implications with the speed of comparisons:

r0 = range(1, 1000000)    
r1 = range(1, 1000000)

l0 = list(r0)    
l1 = list(r1)

Ranges compares super fast:

%timeit r0 == r1
The slowest run took 28.82 times longer than the fastest. This could mean that an intermediate result is being cached 
10000000 loops, best of 3: 160 ns per loop

on the other hand, the lists..

%timeit l0 == l1
10 loops, best of 3: 27.8 ms per loop

Yeah..

As @SuperBiasedMan noted, this only applies to the range objects in Python 3. Python 2 range() is a plain ol' function that returns a list while the 2.x xrange object doesn't have the comparing capabilies (and not only these..) that range objects have in Python 3.

Look at @ajcr's answer for quotes directly from the source code on Python 3 range objects. It's documented in there what the comparison between two different ranges actually entails: Simple quick operations. The range_equals function is utilized in the range_richcompare function for EQ and NE cases and assigned to the tp_richcompare slot for PyRange_Type types.

I believe the implementation of range_equals is pretty readable (because it is nice as simple) to add here:

/* r0 and r1 are pointers to rangeobjects */

/* Check if pointers point to same object, example:    
       >>> r1 = r2 = range(0, 10)
       >>> r1 == r2
   obviously returns True. */
if (r0 == r1)
    return 1;

/* Compare the length of the ranges, if they are equal 
   the checks continue. If they are not, False is returned. */
cmp_result = PyObject_RichCompareBool(r0->length, r1->length, Py_EQ);
/* Return False or error to the caller
       >>> range(0, 10) == range(0, 10, 2)  
   fails here */
if (cmp_result != 1)
    return cmp_result;

/* See if the range has a lenght (non-empty). If the length is 0
   then due to to previous check, the length of the other range is 
   equal to 0. They are equal. */
cmp_result = PyObject_Not(r0->length);
/* Return True or error to the caller. 
       >>> range(0) == range(2, 2, 2)  # True
   (True) gets caught here. Lengths are both zero. */
if (cmp_result != 0)
    return cmp_result;

/* Compare the start values for the ranges, if they don't match
   then we're not dealing with equal ranges. */
cmp_result = PyObject_RichCompareBool(r0->start, r1->start, Py_EQ);
/* Return False or error to the caller. 
   lens are equal, this checks their starting values
       >>> range(0, 10) == range(10, 20)  # False
   Lengths are equal and non-zero, steps don't match.*/
if (cmp_result != 1)
    return cmp_result;

/* Check if the length is equal to 1. 
   If start is the same and length is 1, they represent the same sequence:
       >>> range(0, 10, 10) == range(0, 20, 20)  # True */
one = PyLong_FromLong(1);
if (!one)
    return -1;
cmp_result = PyObject_RichCompareBool(r0->length, one, Py_EQ);
Py_DECREF(one);
/* Return True or error to the caller. */
if (cmp_result != 0)
    return cmp_result;

/* Finally, just compare their steps */
return PyObject_RichCompareBool(r0->step, r1->step, Py_EQ);

I've also scattered some of my own comments here; look at @ajcr's answer for the Python equivalent.

In Python 3.x ,

range(0,3) returns a class of immutable iterable objects that lets you iterate over them, it does not produce lists, and they do not store all the elements in the range in memory, instead they produce the elements on the fly (as you are iterating over them) , whereas list(range(0,3)) produces a list (by iterating over all the elements and appending to the list internally) .

Example -

>>> range(0,3)
range(0, 3)
>>> list(range(0,3))
[0, 1, 2]

Ideally, if you only want to iterate over that range of values , range(0,3) would be faster than (list(range(0,3)) because the latter has the overhead of producing a list before you start iterating over it.

In Python 2.x , range(0,3) produces an list, instead we also had an xrange() function that has similar behavior of range() function from Python 3.x (xrange was renamed to range in Python 3.x)

For Python 3.5, From the documentation -

Range objects implement the collections.abc.Sequence ABC, and provide features such as containment tests, element index lookup, slicing and support for negative indices

So you can do things like -

>>> range(0,10)[5]
5
>>> range(0,10)[3:7]
range(3, 7)
>>> 5 in range(6,10)
False
>>> 7 in range(1,8)
True

And all of these are constant time operations , as can be seen from this test -

In [11]: %timeit a = xrange(0,1000000)[1000]
1000000 loops, best of 3: 342 ns per loop

In [12]: %timeit a = xrange(0,1000000)[10000]
1000000 loops, best of 3: 342 ns per loop

In [13]: %timeit a = xrange(0,1000000)[100000]
1000000 loops, best of 3: 342 ns per loop

In [14]: %timeit a = xrange(0,1000000)[999999]
1000000 loops, best of 3: 342 ns per loop

In [15]: %timeit a = xrange(0,10000000)[9999999]
1000000 loops, best of 3: 339 ns per loop

In [16]: %timeit a = xrange(0,1000000000000)[9999999999]
1000000 loops, best of 3: 341 ns per loop

A "for loop" in most, if not all, programming languages is a mechanism to run a piece of code more than once.

This code:

for i in range(5):
    print i

can be thought of working like this:

i = 0
print i
i = 1
print i
i = 2
print i
i = 3
print i
i = 4
print i

So you see, what happens is not that i gets the value 0, 1, 2, 3, 4 at the same time, but rather sequentially.

I assume that when you say "call a, it gives only 5", you mean like this:

for i in range(5):
    a=i+1
print a

this will print the last value that a was given. Every time the loop iterates, the statement a=i+1 will overwrite the last value a had with the new value.

Code basically runs sequentially, from top to bottom, and a for loop is a way to make the code go back and something again, with a different value for one of the variables.

I hope this answered your question.

Rather than using a decimal step directly, it's much safer to express this in terms of how many points you want. Otherwise, floating-point rounding error is likely to give you a wrong result.

Use the linspace function from the NumPy library (which isn't part of the standard library but is relatively easy to obtain). linspace takes a number of points to return, and also lets you specify whether or not to include the right endpoint:

>>> np.linspace(0,1,11)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9,  1. ])
>>> np.linspace(0,1,10,endpoint=False)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

If you really want to use a floating-point step value, use numpy.arange:

>>> import numpy as np
>>> np.arange(0.0, 1.0, 0.1)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

Floating-point rounding error will cause problems, though. Here's a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers:

>>> numpy.arange(1, 1.3, 0.1)
array([1. , 1.1, 1.2, 1.3])

well, from the help:

>>> help(range)
range(...)
    range([start,] stop[, step]) -> list of integers

    Return a list containing an arithmetic progression of integers.
    range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0.
    When step is given, it specifies the increment (or decrement).
    For example, range(4) returns [0, 1, 2, 3].  The end point is omitted!
    These are exactly the valid indices for a list of 4 elements.

so the last increment is not stop, but the last step before stop.

• in countMe shouldn't the code go up till 18 ;
• why is the last number printed in countMe 15, and not 18 ;
• why is that in the second function oddsOut the function only founts till 7 for j and not 8 even though j is 8 ;
• why is the last number printed in oddsOut 14.

more generally speaking the answer to those questions is that in most of the languages, a range is defined as [start:stop[, i.e. the last value of the range is never included, and the indexes start always at 0. The mess being that in a few languages and when working on algorithmics, ranges start at 1 and are inclusive with the last value.

In the end, if you want to include the last value you can do:

def closed_range(start, stop, step=1):
    return range(start, stop+1, step)

or in your example:

>>> def countMe(num):
>>>     for i in range(0, num+1, 3):
>>>         print (i)
>>> 
>>> countMe(18)
0
3
6
9
12
15
18
>>> 

This is the results from using timeit on the two modules

bhargav@bhargav:~$ python -m timeit "for i in range(10, -1, -1):(i)"
1000000 loops, best of 3: 0.466 usec per loop
bhargav@bhargav:~$ python -m timeit "for i in reversed(range(0, 10 + 1)):(i)"
1000000 loops, best of 3: 0.531 usec per loop

As you can see, the second way is slower that is because it has an additional call to the function reversed.

range takes three arguments, start, stop, and step.

range(20,-1,0):

says to go from 20 to -1 in steps of size 0.

That doesn't make sense, and so python threw an error.

You can't use range for this. The third argument for range is a fixed step used to increment the value: it can't be used to multiply or divide the previous value.

Instead, use a while loop:

i = 1
while i < 100:
   ...
   i *= 10  # or i /= 10 depending on what you want.

If you really want to use a list comprehension, you can do as suggested by Albin Paul:

li = [10**x for x in range(3) ]  # [1, 10, 100]
li = [10**x for x in range(2, -1, -1)  # [100, 10, 1]

‍range(a, b, s) is [a, a+s, a+2*s..., x] where x < b.

So range(0) -> range(0,0,1) generates an empty list. This means the inside of the for loop is skipped which returns True.

You can use any of these:

# Create a range that does not contain 50
for i in [x for x in xrange(100) if x != 50]:
    print i

# Create 2 ranges [0,49] and [51, 100] (Python 2)
for i in range(50) + range(51, 100):
    print i

# Create a iterator and skip 50
xr = iter(xrange(100))
for i in xr:
    print i
    if i == 49:
        next(xr)

# Simply continue in the loop if the number is 50
for i in range(100):
    if i == 50:
        continue
    print i