Mixing
Python Linspace passing in function return only one number
There is two case, the first one, my function Y1 return always the same number. In that case, it doesn't work, y is equal to a integer of 10 and not a array of one thousand 10. The second case, where it's return differents numbers, it works!
First case (Doesn't work as expected)
def Y1(x, N):
return 10
x= np.linspace(-2,2,1000)
y= Y1(x,0) #In that case, it should create a array with 1000 numbers, but it only return one int, 10.
y value: 10 #when it should be [10 10 10 10 10 10...]
The othercases (Does work as expected)
def Y1(x, N):
return x**2
x= np.linspace(-2,2,1000)
y= Y1(x,0) #it returns a array, all numbers are differents
y value:
[4.00000000e+00 3.98400002e+00 3.96803210e+00 3.95209624e+00
3.93619245e+00 3.92032072e+00 3.90448106e+00 3.88867346e+00
3.87289792e+00 3.85715445e+00 3.84144304e+00 3.82576370e+00
3.81011642e+00 3.79450121e+00 3.77891806e+00 3.76336697e+00
...]
Thank you!
python numpy
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
add a comment |
There is two case, the first one, my function Y1 return always the same number. In that case, it doesn't work, y is equal to a integer of 10 and not a array of one thousand 10. The second case, where it's return differents numbers, it works!
First case (Doesn't work as expected)
def Y1(x, N):
return 10
x= np.linspace(-2,2,1000)
y= Y1(x,0) #In that case, it should create a array with 1000 numbers, but it only return one int, 10.
y value: 10 #when it should be [10 10 10 10 10 10...]
The othercases (Does work as expected)
def Y1(x, N):
return x**2
x= np.linspace(-2,2,1000)
y= Y1(x,0) #it returns a array, all numbers are differents
y value:
[4.00000000e+00 3.98400002e+00 3.96803210e+00 3.95209624e+00
3.93619245e+00 3.92032072e+00 3.90448106e+00 3.88867346e+00
3.87289792e+00 3.85715445e+00 3.84144304e+00 3.82576370e+00
3.81011642e+00 3.79450121e+00 3.77891806e+00 3.76336697e+00
...]
Thank you!
python numpy
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
3
Um, why do you expect an array to be returned fromY1when you are only returning an integer?
– R. Kap
Nov 20 '18 at 23:22
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59
add a comment |
There is two case, the first one, my function Y1 return always the same number. In that case, it doesn't work, y is equal to a integer of 10 and not a array of one thousand 10. The second case, where it's return differents numbers, it works!
First case (Doesn't work as expected)
def Y1(x, N):
return 10
x= np.linspace(-2,2,1000)
y= Y1(x,0) #In that case, it should create a array with 1000 numbers, but it only return one int, 10.
y value: 10 #when it should be [10 10 10 10 10 10...]
The othercases (Does work as expected)
def Y1(x, N):
return x**2
x= np.linspace(-2,2,1000)
y= Y1(x,0) #it returns a array, all numbers are differents
y value:
[4.00000000e+00 3.98400002e+00 3.96803210e+00 3.95209624e+00
3.93619245e+00 3.92032072e+00 3.90448106e+00 3.88867346e+00
3.87289792e+00 3.85715445e+00 3.84144304e+00 3.82576370e+00
3.81011642e+00 3.79450121e+00 3.77891806e+00 3.76336697e+00
...]
Thank you!
python numpy
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
There is two case, the first one, my function Y1 return always the same number. In that case, it doesn't work, y is equal to a integer of 10 and not a array of one thousand 10. The second case, where it's return differents numbers, it works!
First case (Doesn't work as expected)
def Y1(x, N):
return 10
x= np.linspace(-2,2,1000)
y= Y1(x,0) #In that case, it should create a array with 1000 numbers, but it only return one int, 10.
y value: 10 #when it should be [10 10 10 10 10 10...]
The othercases (Does work as expected)
def Y1(x, N):
return x**2
x= np.linspace(-2,2,1000)
y= Y1(x,0) #it returns a array, all numbers are differents
y value:
[4.00000000e+00 3.98400002e+00 3.96803210e+00 3.95209624e+00
3.93619245e+00 3.92032072e+00 3.90448106e+00 3.88867346e+00
3.87289792e+00 3.85715445e+00 3.84144304e+00 3.82576370e+00
3.81011642e+00 3.79450121e+00 3.77891806e+00 3.76336697e+00
...]
Thank you!
python numpy
python numpy
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
edited Nov 20 '18 at 23:47
Marc Bourque
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
asked Nov 20 '18 at 23:17
Marc BourqueMarc Bourque
479
479
3
Um, why do you expect an array to be returned fromY1when you are only returning an integer?
– R. Kap
Nov 20 '18 at 23:22
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59
add a comment |
3
Um, why do you expect an array to be returned fromY1when you are only returning an integer?
– R. Kap
Nov 20 '18 at 23:22
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59
3
3
Um, why do you expect an array to be returned from
Y1 when you are only returning an integer?– R. Kap
Nov 20 '18 at 23:22
Um, why do you expect an array to be returned from
Y1 when you are only returning an integer?– R. Kap
Nov 20 '18 at 23:22
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59
add a comment |
1 Answer
1
active
oldest
votes
In the first case, you are returning the constant 10.
In the second case, you are applying the operator **2 to a np array, which overloads ** and applies the power operation element-wise to the full array. This behaviour is known as broadcasting or vectorized aritmethic operations.
This broadcasting overload of the aritmethic methods allows behaviours as the following:
np.array([1, 2, 3, 4]) + 1
>>> array([2, 3, 4, 5])
Which is what your second case uses and your first one does not.
You can learn more about this topic, for example, here.
If you want an array of shape 1000 full of 10s use numpy.full instead:
import numpy as np
y = np.full(1000, 10)
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In the first case, you are returning the constant 10.
In the second case, you are applying the operator **2 to a np array, which overloads ** and applies the power operation element-wise to the full array. This behaviour is known as broadcasting or vectorized aritmethic operations.
This broadcasting overload of the aritmethic methods allows behaviours as the following:
np.array([1, 2, 3, 4]) + 1
>>> array([2, 3, 4, 5])
Which is what your second case uses and your first one does not.
You can learn more about this topic, for example, here.
If you want an array of shape 1000 full of 10s use numpy.full instead:
import numpy as np
y = np.full(1000, 10)
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
add a comment |
In the first case, you are returning the constant 10.
In the second case, you are applying the operator **2 to a np array, which overloads ** and applies the power operation element-wise to the full array. This behaviour is known as broadcasting or vectorized aritmethic operations.
This broadcasting overload of the aritmethic methods allows behaviours as the following:
np.array([1, 2, 3, 4]) + 1
>>> array([2, 3, 4, 5])
Which is what your second case uses and your first one does not.
You can learn more about this topic, for example, here.
If you want an array of shape 1000 full of 10s use numpy.full instead:
import numpy as np
y = np.full(1000, 10)
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
add a comment |
In the first case, you are returning the constant 10.
In the second case, you are applying the operator **2 to a np array, which overloads ** and applies the power operation element-wise to the full array. This behaviour is known as broadcasting or vectorized aritmethic operations.
This broadcasting overload of the aritmethic methods allows behaviours as the following:
np.array([1, 2, 3, 4]) + 1
>>> array([2, 3, 4, 5])
Which is what your second case uses and your first one does not.
You can learn more about this topic, for example, here.
If you want an array of shape 1000 full of 10s use numpy.full instead:
import numpy as np
y = np.full(1000, 10)
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
In the first case, you are returning the constant 10.
In the second case, you are applying the operator **2 to a np array, which overloads ** and applies the power operation element-wise to the full array. This behaviour is known as broadcasting or vectorized aritmethic operations.
This broadcasting overload of the aritmethic methods allows behaviours as the following:
np.array([1, 2, 3, 4]) + 1
>>> array([2, 3, 4, 5])
Which is what your second case uses and your first one does not.
You can learn more about this topic, for example, here.
If you want an array of shape 1000 full of 10s use numpy.full instead:
import numpy as np
y = np.full(1000, 10)
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
answered Nov 20 '18 at 23:46
Julian PellerJulian Peller
8941511
8941511
add a comment |
add a comment |
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3
Um, why do you expect an array to be returned from
Y1when you are only returning an integer?– R. Kap
Nov 20 '18 at 23:22
I added the values generated for both case.
– Marc Bourque
Nov 20 '18 at 23:31
You seriously believe that passing a numpy array as a parameter, that you don't even access from within the function, is somehow going to affect the function's result?
– jasonharper
Nov 20 '18 at 23:46
Sorry, I thought he was calling the function for each element of the array and not that he was passing the array in the parameter. I didn't know that you can make a exposant to an array without a loop. First time using python.
– Marc Bourque
Nov 20 '18 at 23:59