a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
b = []
for i in range(a):
if a[i] == 0.00:
b.append(0)
else:
b.append(1)
What I want is to fill b with 0 and 1 according to if the same index in a is equal to or different from zero. Why am I getting the following error: "TypeError: 'list' object cannot be interpreted as an integer"?
range() takes only integers (start:stop:step). Not a list. So I think you want to use len(list)
a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
b = []
for i in range(len(a)):
if a[i] == 0.00:
b.append(0)
else:
b.append(1)
Lastly, all this can be done using list comprehension
#Note: when iterating through a list, don't use range()
b = [i if i == 0.00 else 1 for i in [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]]
output
[1, 1, 0.0, 1, 1, 0.0]
The bug has already been covered in the other answers. Here's an alternative implementation:
>>> a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
>>> b = [int(bool(x)) for x in a]
>>> b
[1, 1, 0, 1, 1, 0]
As others have said, range takes an integer, not a list. But since no one is covering the biggest flaw in the code, here's a useful warning:
Checking for equality between floating-point numbers are most likely wrong due to the imprecision of floating-point numbers!
As an example, 0.1 + 0.2 == 0.3 is False. Just try it yourself. Instead, check with a small epsilon.
EPSILON = 0.0001
a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
b = []
for i in range(len(a)):
if -EPSILON <= a[i] && a[i] <= EPSILON:
b.append(0)
else:
b.append(1)
Or use a list comprehension, or any other technique used in the other answers.
First of all, as @BuddyBob said, range takes an integer, so that way won't work. Second, I am assuming that you want to loop through the items in a, so it would be easier to use for i in a: like this.
a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
b = []
for i in a:
if i == 0.00:
b.append(0)
else:
b.append(1)
You don't need to use range in your given example. You can just loop through list items:
a = [0.30, 0.15, 0.0, 0.25, 0.30, 0.0]
b = []
for i in a:
b.append(0 if i == 0 else 1)
