How to create in one line a null vector of size 10 but the fifth value being 1 using numpy
Question:
I am able to do it in two lines for the numpy module:
x=np.zeros(10)
x[4]=1
However, I was wondering if its possible to combine the two together
Answers:
x = numpy.array([0,0,0,0,1,0,0,0,0,0])
😛
There are multiple ways to do this. For example, np.arange(10) == 4 gives you an array of all False values except for one True at position 4.
Under the covers, NumPy’s bool values are just 0 and 1 as uint8 (just like Python’s bool values are 0 and 1, although of a unique integral type), so you can just use it as-is in any expression:
>>> np.arange(10) == 4
array([False, False, False, False, True, False, False, False, False, False], dtype=bool)
>>> np.arange(10) * 1
array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0])
>>> np.arange(10) + 23
array([23, 23, 23, 23, 24, 23, 23, 23, 23, 23])
… or view it as uint8 instead of bool:
>>> (np.arange(10) == 4).view(np.uint8)
array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0], dtype=uint8)
… or, if you want normal int values, you can convert it:
>>> (np.arange(10) == 4).astype(int)
  array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0])
And so on.
However, this seems a lot less readable, and it’s also about 20% slower in a quick test, so unless you’re doing this for code-golfing reasons, why?
I don’t that it is possible to combine:
x=np.zeros(10) #1 create an array of values
x[4]=1 #2 assign at least one value of an array a different value
in one line.
Naive:
x = np.zeros(10)[4] = 1 # Fails
fails as value of x is 1 due how python handles chained assignment. Both x and element 4 in array of zeros are assigned value 1.
Therefore, we need to first create an array of zeros, and then assign element assign element 4 a value of 1, and these two cannot be done in one line.
If you need to do this to multiple elements:
my_vect = np.zeros(42)
# set index 1 and 3 to value 1
my_vect[np.array([1,3])] = 1
array = numpy.eye( array_size )[ element_in_array_which_should_be_1 - 1 ]
So to create a null vector of size 10 but the fifth value being 1 in one line is
array = numpy.eye( 10 ) [ 5 - 1 ]
===> array([0., 0., 0., 0., 1., 0., 0., 0., 0., 0.])
🙂
b=np.array([int(x==4) for x in range(10)])
print(b)
[0 0 0 0 1 0 0 0 0 0]
Use this
np.where(np.arange(10)==4,1,0)
I am able to do it in two lines for the numpy module:
x=np.zeros(10)
x[4]=1
However, I was wondering if its possible to combine the two together
x = numpy.array([0,0,0,0,1,0,0,0,0,0])
😛
There are multiple ways to do this. For example, np.arange(10) == 4 gives you an array of all False values except for one True at position 4.
Under the covers, NumPy’s bool values are just 0 and 1 as uint8 (just like Python’s bool values are 0 and 1, although of a unique integral type), so you can just use it as-is in any expression:
>>> np.arange(10) == 4
array([False, False, False, False, True, False, False, False, False, False], dtype=bool)
>>> np.arange(10) * 1
array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0])
>>> np.arange(10) + 23
array([23, 23, 23, 23, 24, 23, 23, 23, 23, 23])
… or view it as uint8 instead of bool:
>>> (np.arange(10) == 4).view(np.uint8)
array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0], dtype=uint8)
… or, if you want normal int values, you can convert it:
>>> (np.arange(10) == 4).astype(int)
  array([0, 0, 0, 0, 1, 0, 0, 0, 0, 0])
And so on.
However, this seems a lot less readable, and it’s also about 20% slower in a quick test, so unless you’re doing this for code-golfing reasons, why?
I don’t that it is possible to combine:
x=np.zeros(10) #1 create an array of values
x[4]=1 #2 assign at least one value of an array a different value
in one line.
Naive:
x = np.zeros(10)[4] = 1 # Fails
fails as value of x is 1 due how python handles chained assignment. Both x and element 4 in array of zeros are assigned value 1.
Therefore, we need to first create an array of zeros, and then assign element assign element 4 a value of 1, and these two cannot be done in one line.
If you need to do this to multiple elements:
my_vect = np.zeros(42)
# set index 1 and 3 to value 1
my_vect[np.array([1,3])] = 1
array = numpy.eye( array_size )[ element_in_array_which_should_be_1 - 1 ]
So to create a null vector of size 10 but the fifth value being 1 in one line is
array = numpy.eye( 10 ) [ 5 - 1 ]
===> array([0., 0., 0., 0., 1., 0., 0., 0., 0., 0.])
🙂
b=np.array([int(x==4) for x in range(10)])
print(b)
[0 0 0 0 1 0 0 0 0 0]
Use this
np.where(np.arange(10)==4,1,0)