本文整理汇总了Python中numpy.core.numerictypes.issubdtype函数的典型用法代码示例。如果您正苦于以下问题:Python issubdtype函数的具体用法?Python issubdtype怎么用?Python issubdtype使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了issubdtype函数的4个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: ix_
def ix_(*args):
"""
Construct an open mesh from multiple sequences.
This function takes N 1-D sequences and returns N outputs with N
dimensions each, such that the shape is 1 in all but one dimension
and the dimension with the non-unit shape value cycles through all
N dimensions.
Using `ix_` one can quickly construct index arrays that will index
the cross product. ``a[np.ix_([1,3],[2,5])]`` returns the array
``[[a[1,2] a[1,5]], [a[3,2] a[3,5]]]``.
Parameters
----------
args : 1-D sequences
Returns
-------
out : tuple of ndarrays
N arrays with N dimensions each, with N the number of input
sequences. Together these arrays form an open mesh.
See Also
--------
ogrid, mgrid, meshgrid
Examples
--------
>>> a = np.arange(10).reshape(2, 5)
>>> a
array([[0, 1, 2, 3, 4],
[5, 6, 7, 8, 9]])
>>> ixgrid = np.ix_([0,1], [2,4])
>>> ixgrid
(array([[0],
[1]]), array([[2, 4]]))
>>> ixgrid[0].shape, ixgrid[1].shape
((2, 1), (1, 2))
>>> a[ixgrid]
array([[2, 4],
[7, 9]])
"""
out = []
nd = len(args)
for k, new in enumerate(args):
new = asarray(new)
if new.ndim != 1:
raise ValueError("Cross index must be 1 dimensional")
if new.size == 0:
# Explicitly type empty arrays to avoid float default
new = new.astype(_nx.intp)
if issubdtype(new.dtype, _nx.bool_):
new, = new.nonzero()
new = new.reshape((1,)*k + (new.size,) + (1,)*(nd-k-1))
out.append(new)
return tuple(out)
开发者ID:8ballbb,项目名称:ProjectRothar,代码行数:58,代码来源:index_tricks.py
示例2: compare
def compare(x, y):
try:
if npany(gisinf(x)) or npany( gisinf(y)):
xinfid = gisinf(x)
yinfid = gisinf(y)
if not xinfid == yinfid:
return False
# if one item, x and y is +- inf
if x.size == y.size == 1:
return x == y
x = x[~xinfid]
y = y[~yinfid]
except TypeError:
pass
z = abs(x-y)
if not issubdtype(z.dtype, number):
z = z.astype(float_) # handle object arrays
return around(z, decimal) <= 10.0**(-decimal)
开发者ID:chadnetzer,项目名称:numpy-gaurdro,代码行数:18,代码来源:utils.py
示例3: matrix_power
def matrix_power(M, n):
"""
Raise a square matrix to the (integer) power `n`.
For positive integers `n`, the power is computed by repeated matrix
squarings and matrix multiplications. If ``n == 0``, the identity matrix
of the same shape as M is returned. If ``n < 0``, the inverse
is computed and then raised to the ``abs(n)``.
Parameters
----------
M : ndarray or matrix object
Matrix to be "powered." Must be square, i.e. ``M.shape == (m, m)``,
with `m` a positive integer.
n : int
The exponent can be any integer or long integer, positive,
negative, or zero.
Returns
-------
M**n : ndarray or matrix object
The return value is the same shape and type as `M`;
if the exponent is positive or zero then the type of the
elements is the same as those of `M`. If the exponent is
negative the elements are floating-point.
Raises
------
LinAlgError
If the matrix is not numerically invertible.
See Also
--------
matrix
Provides an equivalent function as the exponentiation operator
(``**``, not ``^``).
Examples
--------
>>> from numpy import linalg as LA
>>> i = np.array([[0, 1], [-1, 0]]) # matrix equiv. of the imaginary unit
>>> LA.matrix_power(i, 3) # should = -i
array([[ 0, -1],
[ 1, 0]])
>>> LA.matrix_power(np.matrix(i), 3) # matrix arg returns matrix
matrix([[ 0, -1],
[ 1, 0]])
>>> LA.matrix_power(i, 0)
array([[1, 0],
[0, 1]])
>>> LA.matrix_power(i, -3) # should = 1/(-i) = i, but w/ f.p. elements
array([[ 0., 1.],
[-1., 0.]])
Somewhat more sophisticated example
>>> q = np.zeros((4, 4))
>>> q[0:2, 0:2] = -i
>>> q[2:4, 2:4] = i
>>> q # one of the three quarternion units not equal to 1
array([[ 0., -1., 0., 0.],
[ 1., 0., 0., 0.],
[ 0., 0., 0., 1.],
[ 0., 0., -1., 0.]])
>>> LA.matrix_power(q, 2) # = -np.eye(4)
array([[-1., 0., 0., 0.],
[ 0., -1., 0., 0.],
[ 0., 0., -1., 0.],
[ 0., 0., 0., -1.]])
"""
M = asanyarray(M)
if len(M.shape) != 2 or M.shape[0] != M.shape[1]:
raise ValueError("input must be a square array")
if not issubdtype(type(n), int):
raise TypeError("exponent must be an integer")
from numpy.linalg import inv
if n==0:
M = M.copy()
M[:] = identity(M.shape[0])
return M
elif n<0:
M = inv(M)
n *= -1
result = M
if n <= 3:
for _ in range(n-1):
result=N.dot(result, M)
return result
# binary decomposition to reduce the number of Matrix
# multiplications for n > 3.
beta = binary_repr(n)
Z, q, t = M, 0, len(beta)
while beta[t-q-1] == '0':
Z = N.dot(Z, Z)
q += 1
#.........这里部分代码省略.........
开发者ID:ihuston,项目名称:numpy,代码行数:101,代码来源:defmatrix.py
示例4: matrix_power
def matrix_power(M,n):
"""
Raise a square matrix to the (integer) power n.
For positive integers n, the power is computed by repeated matrix
squarings and matrix multiplications. If n=0, the identity matrix
of the same type as M is returned. If n<0, the inverse is computed
and raised to the exponent.
Parameters
----------
M : array_like
Must be a square array (that is, of dimension two and with
equal sizes).
n : integer
The exponent can be any integer or long integer, positive
negative or zero.
Returns
-------
M to the power n
The return value is a an array the same shape and size as M;
if the exponent was positive or zero then the type of the
elements is the same as those of M. If the exponent was negative
the elements are floating-point.
Raises
------
LinAlgException
If the matrix is not numerically invertible, an exception is raised.
See Also
--------
The matrix() class provides an equivalent function as the exponentiation
operator.
Examples
--------
>>> np.linalg.matrix_power(np.array([[0,1],[-1,0]]),10)
array([[-1, 0],
[ 0, -1]])
"""
M = asanyarray(M)
if len(M.shape) != 2 or M.shape[0] != M.shape[1]:
raise ValueError("input must be a square array")
if not issubdtype(type(n),int):
raise TypeError("exponent must be an integer")
from numpy.linalg import inv
if n==0:
M = M.copy()
M[:] = identity(M.shape[0])
return M
elif n<0:
M = inv(M)
n *= -1
result = M
if n <= 3:
for _ in range(n-1):
result=N.dot(result,M)
return result
# binary decomposition to reduce the number of Matrix
# multiplications for n > 3.
beta = binary_repr(n)
Z,q,t = M,0,len(beta)
while beta[t-q-1] == '0':
Z = N.dot(Z,Z)
q += 1
result = Z
for k in range(q+1,t):
Z = N.dot(Z,Z)
if beta[t-k-1] == '1':
result = N.dot(result,Z)
return result
开发者ID:chadnetzer,项目名称:numpy-gaurdro,代码行数:78,代码来源:defmatrix.py
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