Python numeric.where函数代码示例

    def __train__(self, data, labels):
        l = labels.reshape((-1,1))
        self.__trainingData__ = data
        self.__trainingLabels__ = l
        N = len(l)
        H = zeros((N,N))
        for i in range(N):
            for j in range(N):
                H[i,j] = self.__trainingLabels__[i]*self.__trainingLabels__[j]*self.__kernelFunc__(self.__trainingData__[i],self.__trainingData__[j])
        f = -1.0*ones(labels.shape)
        lb = zeros(labels.shape)
        ub = self.C * ones(labels.shape)
        Aeq = labels
        beq = 0.0
        suppressOut = True
        if suppressOut:
            devnull = open('/dev/null', 'w')
            oldstdout_fno = os.dup(sys.stdout.fileno())
            os.dup2(devnull.fileno(), 1)
        p = QP(matrix(H),f.tolist(),lb=lb.tolist(),ub=ub.tolist(),Aeq=Aeq.tolist(),beq=beq)
        r = p.solve('cvxopt_qp')
        if suppressOut:
            os.dup2(oldstdout_fno, 1)
        lim = 1e-4
        r.xf[where(abs(r.xf)<lim)] = 0
        self.__lambdas__ = r.xf
        nonzeroindexes = where(r.xf>lim)[0]
#        l1 = nonzeroindexes[0]
#        self.w0 = 1.0/labels[l1]-dot(self.w,data[l1])
        self.numSupportVectors = len(nonzeroindexes)

def triu(m, k=0):
    """
    Upper triangle of an array.

    Return a copy of a matrix with the elements below the `k`-th diagonal
    zeroed.

    Please refer to the documentation for `tril` for further details.

    See Also
    --------
    tril : lower triangle of an array

    Examples
    --------
    >>> np.triu([[1,2,3],[4,5,6],[7,8,9],[10,11,12]], -1)
    array([[ 1,  2,  3],
           [ 4,  5,  6],
           [ 0,  8,  9],
           [ 0,  0, 12]])

    """
    m = asanyarray(m)
    mask = tri(*m.shape[-2:], k=k-1, dtype=bool)

    return where(mask, zeros(1, m.dtype), m)

def calcN(classKernels, trainLabels):
    N = zeros((len(trainLabels), len(trainLabels)))
    for i, l in enumerate(unique(trainLabels)):
        numExamplesWithLabel = len(where(trainLabels == l)[0])
        Idiff = identity(numExamplesWithLabel, Float64) - (1.0 / numExamplesWithLabel) * ones(numExamplesWithLabel, Float64)
        firstDot = dot(classKernels[i], Idiff)
        labelTerm = dot(firstDot, transpose(classKernels[i]))
        N += labelTerm
    N = nan_to_num(N)
    #make N more numerically stable
    #if I had more time, I would train this parameter, but I don't
    additionToN = ((mean(diag(N)) + 1) / 100.0) * identity(N.shape[0], Float64) 
    N += additionToN
            
    #make sure N is invertable
    for i in range(1000):
        try:
            inv(N)
        except LinAlgError:
            #doing this to make sure the maxtrix is invertable
            #large value supported by section titled
            #"numerical issues and regularization" in the paper
            N += additionToN

    return N

 def train(self, data, labels):
     l = labels.reshape((-1,1))
     xy = data * l
     H = dot(xy,transpose(xy))
     f = -1.0*ones(labels.shape)
     lb = zeros(labels.shape)
     ub = self.C * ones(labels.shape)
     Aeq = labels
     beq = 0.0
     p = QP(matrix(H),f.tolist(),lb=lb.tolist(),ub=ub.tolist(),Aeq=Aeq.tolist(),beq=beq)
     r = p.solve('cvxopt_qp')
     r.xf[where(r.xf<1e-3)] = 0
     self.w = dot(r.xf*labels,data)
     nonzeroindexes = where(r.xf>1e-4)[0]
     l1 = nonzeroindexes[0]
     self.w0 = 1.0/labels[l1]-dot(self.w,data[l1])
     self.numSupportVectors = len(nonzeroindexes)

def bartlett(M):
    """bartlett(M) returns the M-point Bartlett window.
    """
    if M < 1:
        return array([])
    if M == 1:
        return ones(1, float)
    n = arange(0,M)
    return where(less_equal(n,(M-1)/2.0),2.0*n/(M-1),2.0-2.0*n/(M-1))

def getClassKernels(fullKernelMatrix, trainLabels):
    #create a matrix where rows correspond to all examples
    #and columns correspond to examples of a specific class
    #so if l is the total number of examples, and lj is the number of examples in class j
    #then we're creating an l x lj matrix
    uniqueLabels = unique(trainLabels)
    ret = []
    for l in uniqueLabels:
        labelIndexes = where(trainLabels == l)[0]
        k = zeros((len(fullKernelMatrix), len(labelIndexes)))
        for r in range(len(k)):
            for c in range(len(k[r])):
                k[r][c] = fullKernelMatrix[r][labelIndexes[c]]
        ret.append(k)
    return ret        

def fix(x, y=None):
    """
    Round to nearest integer towards zero.

    Round an array of floats element-wise to nearest integer towards zero.
    The rounded values are returned as floats.

    Parameters
    ----------
    x : array_like
        An array of floats to be rounded
    y : ndarray, optional
        Output array

    Returns
    -------
    out : ndarray of floats
        The array of rounded numbers

    See Also
    --------
    trunc, floor, ceil
    around : Round to given number of decimals

    Examples
    --------
    >>> np.fix(3.14)
    3.0
    >>> np.fix(3)
    3.0
    >>> np.fix([2.1, 2.9, -2.1, -2.9])
    array([ 2.,  2., -2., -2.])

    """
    x = nx.asanyarray(x)
    y1 = nx.floor(x)
    y2 = nx.ceil(x)
    if y is None:
        y = nx.asanyarray(y1)
    y[...] = nx.where(x >= 0, y1, y2)
    return y

def tril(m, k=0):
    """
    Lower triangle of an array.

    Return a copy of an array with elements above the `k`-th diagonal zeroed.

    Parameters
    ----------
    m : array_like, shape (M, N)
        Input array.
    k : int, optional
        Diagonal above which to zero elements.  `k = 0` (the default) is the
        main diagonal, `k < 0` is below it and `k > 0` is above.

    Returns
    -------
    tril : ndarray, shape (M, N)
        Lower triangle of `m`, of same shape and data-type as `m`.

    See Also
    --------
    triu : same thing, only for the upper triangle

    Examples
    --------
    >>> np.tril([[1,2,3],[4,5,6],[7,8,9],[10,11,12]], -1)
    array([[ 0,  0,  0],
           [ 4,  0,  0],
           [ 7,  8,  0],
           [10, 11, 12]])

    """
    m = asanyarray(m)
    mask = tri(*m.shape[-2:], k=k, dtype=bool)

    return where(mask, m, zeros(1, m.dtype))

def mask_indices(n, mask_func, k=0):
    """
    Return the indices to access (n, n) arrays, given a masking function.

    Assume `mask_func` is a function that, for a square array a of size
    ``(n, n)`` with a possible offset argument `k`, when called as
    ``mask_func(a, k)`` returns a new array with zeros in certain locations
    (functions like `triu` or `tril` do precisely this). Then this function
    returns the indices where the non-zero values would be located.

    Parameters
    ----------
    n : int
        The returned indices will be valid to access arrays of shape (n, n).
    mask_func : callable
        A function whose call signature is similar to that of `triu`, `tril`.
        That is, ``mask_func(x, k)`` returns a boolean array, shaped like `x`.
        `k` is an optional argument to the function.
    k : scalar
        An optional argument which is passed through to `mask_func`. Functions
        like `triu`, `tril` take a second argument that is interpreted as an
        offset.

    Returns
    -------
    indices : tuple of arrays.
        The `n` arrays of indices corresponding to the locations where
        ``mask_func(np.ones((n, n)), k)`` is True.

    See Also
    --------
    triu, tril, triu_indices, tril_indices

    Notes
    -----
    .. versionadded:: 1.4.0

    Examples
    --------
    These are the indices that would allow you to access the upper triangular
    part of any 3x3 array:

    >>> iu = np.mask_indices(3, np.triu)

    For example, if `a` is a 3x3 array:

    >>> a = np.arange(9).reshape(3, 3)
    >>> a
    array([[0, 1, 2],
           [3, 4, 5],
           [6, 7, 8]])
    >>> a[iu]
    array([0, 1, 2, 4, 5, 8])

    An offset can be passed also to the masking function.  This gets us the
    indices starting on the first diagonal right of the main one:

    >>> iu1 = np.mask_indices(3, np.triu, 1)

    with which we now extract only three elements:

    >>> a[iu1]
    array([1, 2, 5])

    """
    m = ones((n,n), int)
    a = mask_func(m, k)
    return where(a != 0)

def triu_indices(n, k=0, m=None):
    """
    Return the indices for the upper-triangle of an (n, m) array.

    Parameters
    ----------
    n : int
        The size of the arrays for which the returned indices will
        be valid.
    k : int, optional
        Diagonal offset (see `triu` for details).
    m : int, optional
        .. versionadded:: 1.9.0

        The column dimension of the arrays for which the returned
        arrays will be valid.
        By default `m` is taken equal to `n`.


    Returns
    -------
    inds : tuple, shape(2) of ndarrays, shape(`n`)
        The indices for the triangle. The returned tuple contains two arrays,
        each with the indices along one dimension of the array.  Can be used
        to slice a ndarray of shape(`n`, `n`).

    See also
    --------
    tril_indices : similar function, for lower-triangular.
    mask_indices : generic function accepting an arbitrary mask function.
    triu, tril

    Notes
    -----
    .. versionadded:: 1.4.0

    Examples
    --------
    Compute two different sets of indices to access 4x4 arrays, one for the
    upper triangular part starting at the main diagonal, and one starting two
    diagonals further right:

    >>> iu1 = np.triu_indices(4)
    >>> iu2 = np.triu_indices(4, 2)

    Here is how they can be used with a sample array:

    >>> a = np.arange(16).reshape(4, 4)
    >>> a
    array([[ 0,  1,  2,  3],
           [ 4,  5,  6,  7],
           [ 8,  9, 10, 11],
           [12, 13, 14, 15]])

    Both for indexing:

    >>> a[iu1]
    array([ 0,  1,  2,  3,  5,  6,  7, 10, 11, 15])

    And for assigning values:

    >>> a[iu1] = -1
    >>> a
    array([[-1, -1, -1, -1],
           [ 4, -1, -1, -1],
           [ 8,  9, -1, -1],
           [12, 13, 14, -1]])

    These cover only a small part of the whole array (two diagonals right
    of the main one):

    >>> a[iu2] = -10
    >>> a
    array([[ -1,  -1, -10, -10],
           [  4,  -1,  -1, -10],
           [  8,   9,  -1,  -1],
           [ 12,  13,  14,  -1]])

    """
    return where(~tri(n, m, k=k - 1, dtype=bool))

def sinhc(x):
    x = np.asanyarray(x)
    y = where(x == 0, 1.0e-20, x)
    return np.sinh(y)/y

#.........这里部分代码省略.........

    nbin = empty(D, int)
    edges = D*[None]
    dedges = D*[None]
    if weights is not None:
        weights = asarray(weights)

    try:
        M = len(bins)
        if M != D:
            raise AttributeError, 'The dimension of bins must be a equal to the dimension of the sample x.'
    except TypeError:
        bins = D*[bins]

    # Select range for each dimension
    # Used only if number of bins is given.
    if range is None:
        smin = atleast_1d(array(sample.min(0), float))
        smax = atleast_1d(array(sample.max(0), float))
    else:
        smin = zeros(D)
        smax = zeros(D)
        for i in arange(D):
            smin[i], smax[i] = range[i]

    # Make sure the bins have a finite width.
    for i in arange(len(smin)):
        if smin[i] == smax[i]:
            smin[i] = smin[i] - .5
            smax[i] = smax[i] + .5

    # Create edge arrays
    for i in arange(D):
        if isscalar(bins[i]):
            nbin[i] = bins[i] + 2 # +2 for outlier bins
            edges[i] = linspace(smin[i], smax[i], nbin[i]-1)
        else:
            edges[i] = asarray(bins[i], float)
            nbin[i] = len(edges[i])+1  # +1 for outlier bins
        dedges[i] = diff(edges[i])

    nbin =  asarray(nbin)

    # Compute the bin number each sample falls into.
    Ncount = {}
    for i in arange(D):
        Ncount[i] = digitize(sample[:,i], edges[i])

    # Using digitize, values that fall on an edge are put in the right bin.
    # For the rightmost bin, we want values equal to the right
    # edge to be counted in the last bin, and not as an outlier.
    outliers = zeros(N, int)
    for i in arange(D):
        # Rounding precision
        decimal = int(-log10(dedges[i].min())) +6
        # Find which points are on the rightmost edge.
        on_edge = where(around(sample[:,i], decimal) == around(edges[i][-1], decimal))[0]
        # Shift these points one bin to the left.
        Ncount[i][on_edge] -= 1

    # Flattened histogram matrix (1D)
    hist = zeros(nbin.prod(), float)

    # Compute the sample indices in the flattened histogram matrix.
    ni = nbin.argsort()
    shape = []
    xy = zeros(N, int)
    for i in arange(0, D-1):
        xy += Ncount[ni[i]] * nbin[ni[i+1:]].prod()
    xy += Ncount[ni[-1]]

    # Compute the number of repetitions in xy and assign it to the flattened histmat.
    if len(xy) == 0:
        return zeros(nbin-2, int), edges

    flatcount = bincount(xy, weights)
    a = arange(len(flatcount))
    hist[a] = flatcount

    # Shape into a proper matrix
    hist = hist.reshape(sort(nbin))
    for i in arange(nbin.size):
        j = ni[i]
        hist = hist.swapaxes(i,j)
        ni[i],ni[j] = ni[j],ni[i]

    # Remove outliers (indices 0 and -1 for each dimension).
    core = D*[slice(1,-1)]
    hist = hist[core]

    # Normalize if normed is True
    if normed:
        s = hist.sum()
        for i in arange(D):
            shape = ones(D, int)
            shape[i] = nbin[i]-2
            hist = hist / dedges[i].reshape(shape)
        hist /= s

    return hist, edges

def sinc(x):
    """sinc(x) returns sin(pi*x)/(pi*x) at all points of array x.
    """
    y = pi* where(x == 0, 1.0e-20, x)
    return sin(y)/y

def mask_indices(n,mask_func,k=0):
    """Return the indices to access (n,n) arrays, given a masking function.

    Assume mask_func() is a function that, for a square array a of size (n,n)
    with a possible offset argument k, when called as mask_func(a,k) returns a
    new array with zeros in certain locations (functions like triu() or tril()
    do precisely this).  Then this function returns the indices where the
    non-zero values would be located.

    Parameters
    ----------
    n : int
      The returned indices will be valid to access arrays of shape (n,n).

    mask_func : callable
      A function whose api is similar to that of numpy.tri{u,l}.  That is,
      mask_func(x,k) returns a boolean array, shaped like x.  k is an optional
      argument to the function.

    k : scalar
      An optional argument which is passed through to mask_func().  Functions
      like tri{u,l} take a second argument that is interpreted as an offset.

    Returns
    -------
    indices : an n-tuple of index arrays.
      The indices corresponding to the locations where mask_func(ones((n,n)),k)
      is True.

    Notes
    -----
    .. versionadded:: 1.4.0

    Examples
    --------
    These are the indices that would allow you to access the upper triangular
    part of any 3x3 array:
    >>> iu = mask_indices(3,np.triu)

    For example, if `a` is a 3x3 array:
    >>> a = np.arange(9).reshape(3,3)
    >>> a
    array([[0, 1, 2],
           [3, 4, 5],
           [6, 7, 8]])

    Then:
    >>> a[iu]
    array([0, 1, 2, 4, 5, 8])

    An offset can be passed also to the masking function.  This gets us the
    indices starting on the first diagonal right of the main one:
    >>> iu1 = mask_indices(3,np.triu,1)

    with which we now extract only three elements:
    >>> a[iu1]
    array([1, 2, 5])
    """ 
    m = ones((n,n),int)
    a = mask_func(m,k)
    return where(a != 0)