**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) `

ID : 274382

viewed : 26

Tags : pythonpython-2.7rangepython

95

**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) `

82

If you have `numpy`

, here are two ways to do it:

`numpy.arange(0, 100, 0.5) numpy.linspace(0, 100, 200, endpoint=False) `

76

You have to use integer steps for `range`

() and `xrange()`

. That's why your 0.5 step gets internally converted to 0 and you get that error. Try `for i in [j / 2.0 for j in xrange(100 * 2)]:`

67

You'll have to either create the loop manually, or define your own custom range function. The built-in requires an integer step value.

59

`for x in map(lambda i: i * 0.5, range(0,200)): #Do something with x `