Reputation: 33
import numpy as np
def Vin(t):
inputs = []
for i in range (1000):
if (-1)**(np.floor( 2 * t[i] )) == 1:
Vin = (1)
inputs.append(Vin)
else:
Vin = (-1)
inputs.append(Vin)
return inputs
when I use this function on a range of t
values, I only get one result,
i.e.
input1=Vin(tpoints)
print (input1)
only gives [1], whereas I want the function to do it for every t
value.
Upvotes: 1
Views: 5197
Reputation: 7091
You are returning from with in the for loop,So you would return from the function on first iteration of the loop.
you may re-writ e the function as following
Code
def Vin(t):
inputs = []
for i in range (1000):
if (-1)**(np.floor( 2 * t[i] )) == 1:
inputs.append(1)
else:
inputs.append(-1)
return inputs
Check the indentation of return inputs in the function here .
BTW, your function can be reduced to as more pythonic and efficient code
def Vin(t):
reduce map(lambda x:int((-1)**(np.floor( t[i] < 1))), range (1000))
Upvotes: 0
Reputation: 140216
As others said, there's an indentation error in your return
statement. I couldn't resist rewriting your code in a more pythonic way, avoiding all that cumbersome loops and the resulting errors:
return [1 if (-1)**(np.floor( 2 * t[i] )) == 1 else -1 for i in range (1000)]
that list comprenhension + ternary construction statement creates your -1,1 array in 1 line, much faster than you'd write with a loop.
Upvotes: 3
Reputation: 14389
try:
for i in range (1000):
if (-1)**(np.floor( 2 * t[i] )) == 1:
Vin = (1)
inputs.append(Vin)
else:
Vin = (-1)
inputs.append(Vin)
return inputs
Current indentation of return , exits the for loop after 1st iteration
Upvotes: 0