NumPy 中文文档

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Less 基础

# Less 基础

# 广播(Broadcasting)规则

Broadcasting允许通用函数以有意义的方式处理具有不完全相同形状的输入。

Broadcasting的第一个规则是,如果所有输入数组不具有相同数量的维度,则“1”将被重复地添加到较小数组的形状,直到所有数组具有相同数量的维度。

Broadcasting的第二个规则确保沿着特定维度具有大小为1的数组表现得好像它们具有沿着该维度具有最大形状的数组的大小。假定数组元素的值沿“Broadcasting”数组的该维度相同。

在应用广播规则之后,所有阵列的大小必须匹配。更多细节可以在 Broadcasting 中找到。

# 花式索引和索引技巧

NumPy提供了比常规Python序列更多的索引能力。正如我们前面看到的,除了通过整数和切片进行索引之外,还可以使用整数数组和布尔数组进行索引。

# 使用索引数组索引

>>> a = np.arange(12)**2                       # the first 12 square numbers
>>> i = np.array( [ 1,1,3,8,5 ] )              # an array of indices
>>> a[i]                                       # the elements of a at the positions i
array([ 1,  1,  9, 64, 25])
>>>
>>> j = np.array( [ [ 3, 4], [ 9, 7 ] ] )      # a bidimensional array of indices
>>> a[j]                                       # the same shape as j
array([[ 9, 16],
       [81, 49]])

当被索引的数组 a 是一个多维数组,单个索引数组指的是 a 的第一个维度。以下示例通过使用调色板将标签图像转换为彩色图像来作为举例。

>>> palette = np.array( [ [0,0,0],                # black
...                       [255,0,0],              # red
...                       [0,255,0],              # green
...                       [0,0,255],              # blue
...                       [255,255,255] ] )       # white
>>> image = np.array( [ [ 0, 1, 2, 0 ],           # each value corresponds to a color in the palette
...                     [ 0, 3, 4, 0 ]  ] )
>>> palette[image]                            # the (2,4,3) color image
array([[[  0,   0,   0],
        [255,   0,   0],
        [  0, 255,   0],
        [  0,   0,   0]],
       [[  0,   0,   0],
        [  0,   0, 255],
        [255, 255, 255],
        [  0,   0,   0]]])

我们也可以给出多个维度的索引。每个维度的索引数组必须具有相同的形状。

>>> a = np.arange(12).reshape(3,4)
>>> a
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
>>> i = np.array( [ [0,1],                        # indices for the first dim of a
...                 [1,2] ] )
>>> j = np.array( [ [2,1],                        # indices for the second dim
...                 [3,3] ] )
>>>
>>> a[i,j]                                     # i and j must have equal shape
array([[ 2,  5],
       [ 7, 11]])
>>>
>>> a[i,2]
array([[ 2,  6],
       [ 6, 10]])
>>>
>>> a[:,j]                                     # i.e., a[ : , j]
array([[[ 2,  1],
        [ 3,  3]],
       [[ 6,  5],
        [ 7,  7]],
       [[10,  9],
        [11, 11]]])

当然,我们可以把 ij 放在一个序列中(比如一个列表),然后用列表进行索引。

>>> l = [i,j]
>>> a[l]                                       # equivalent to a[i,j]
array([[ 2,  5],
       [ 7, 11]])

然而,我们不能将 ij 放入一个数组中,因为这个数组将被解释为索引第一个维度。

>>> s = np.array( [i,j] )
>>> a[s]                                       # not what we want
Traceback (most recent call last):
  File "<stdin>", line 1, in ?
IndexError: index (3) out of range (0<=index<=2) in dimension 0
>>>
>>> a[tuple(s)]                                # same as a[i,j]
array([[ 2,  5],
       [ 7, 11]])

索引数组的另一个常见用途是搜索时间相关序列的最大值:

>>> time = np.linspace(20, 145, 5)                 # time scale
>>> data = np.sin(np.arange(20)).reshape(5,4)      # 4 time-dependent series
>>> time
array([  20.  ,   51.25,   82.5 ,  113.75,  145.  ])
>>> data
array([[ 0.        ,  0.84147098,  0.90929743,  0.14112001],
       [-0.7568025 , -0.95892427, -0.2794155 ,  0.6569866 ],
       [ 0.98935825,  0.41211849, -0.54402111, -0.99999021],
       [-0.53657292,  0.42016704,  0.99060736,  0.65028784],
       [-0.28790332, -0.96139749, -0.75098725,  0.14987721]])
>>>
>>> ind = data.argmax(axis=0)                  # index of the maxima for each series
>>> ind
array([2, 0, 3, 1])
>>>
>>> time_max = time[ind]                       # times corresponding to the maxima
>>>
>>> data_max = data[ind, range(data.shape[1])] # => data[ind[0],0], data[ind[1],1]...
>>>
>>> time_max
array([  82.5 ,   20.  ,  113.75,   51.25])
>>> data_max
array([ 0.98935825,  0.84147098,  0.99060736,  0.6569866 ])
>>>
>>> np.all(data_max == data.max(axis=0))
True

你还可以使用数组索引作为目标来赋值:

>>> a = np.arange(5)
>>> a
array([0, 1, 2, 3, 4])
>>> a[[1,3,4]] = 0
>>> a
array([0, 0, 2, 0, 0])

然而,当索引列表包含重复时,赋值完成多次,留下最后一个值:

>>> a = np.arange(5)
>>> a[[0,0,2]]=[1,2,3]
>>> a
array([2, 1, 3, 3, 4])

这相当合理,但如果你想使用Python的 += 构造要小心,因为这可能得不到你想要的效果:

>>> a = np.arange(5)
>>> a[[0,0,2]]+=1
>>> a
array([1, 1, 3, 3, 4])

即使0在索引列表中出现两次,第0个元素只会增加一次。这是因为Python要求“a + = 1”等同于“a = a + 1”。

# 使用布尔值作为数组索引

当我们用(整数)索引数组索引数组时,我们提供了要选择的索引列表。使用布尔值作为索引时,方法是不同的;我们明确地选择数组中的哪些元素我们想要的,哪些不是。

我们可以想到的布尔索引最自然的方式是使用与原始数组具有相同形状的布尔数组:

>>> a = np.arange(12).reshape(3,4)
>>> b = a > 4
>>> b                                          # b is a boolean with a's shape
array([[False, False, False, False],
       [False,  True,  True,  True],
       [ True,  True,  True,  True]])
>>> a[b]                                       # 1d array with the selected elements
array([ 5,  6,  7,  8,  9, 10, 11])

# 此属性在赋值时非常有用:

    >>> a[b] = 0                                   # All elements of 'a' higher than 4 become 0
    >>> a
    array([[0, 1, 2, 3],
           [4, 0, 0, 0],
           [0, 0, 0, 0]])

你可以查看以下示例,了解如何使用布尔索引生成 Mandelbrot 集的图像:

>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> def mandelbrot( h,w, maxit=20 ):
...     """Returns an image of the Mandelbrot fractal of size (h,w)."""
...     y,x = np.ogrid[ -1.4:1.4:h*1j, -2:0.8:w*1j ]
...     c = x+y*1j
...     z = c
...     divtime = maxit + np.zeros(z.shape, dtype=int)
...
...     for i in range(maxit):
...         z = z**2 + c
...         diverge = z*np.conj(z) > 2**2            # who is diverging
...         div_now = diverge & (divtime==maxit)  # who is diverging now
...         divtime[div_now] = i                  # note when
...         z[diverge] = 2                        # avoid diverging too much
...
...     return divtime
>>> plt.imshow(mandelbrot(400,400))
>>> plt.show()

quickstart-1

第二种使用布尔索引的方法更类似于整数索引;对于数组的每个维度,我们给出一个一维布尔数组,选择我们想要的切片:

>>> a = np.arange(12).reshape(3,4)
>>> b1 = np.array([False,True,True])             # first dim selection
>>> b2 = np.array([True,False,True,False])       # second dim selection
>>>
>>> a[b1,:]                                   # selecting rows
array([[ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
>>>
>>> a[b1]                                     # same thing
array([[ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
>>>
>>> a[:,b2]                                   # selecting columns
array([[ 0,  2],
       [ 4,  6],
       [ 8, 10]])
>>>
>>> a[b1,b2]                                  # a weird thing to do
array([ 4, 10])

请注意,1D布尔数组的长度必须与你要切片的维度(或轴)的长度一致。在前面的示例中, b1 是rank为1的数组,其长度为3( a 中行的数量), b2 (长度4)适合于索引 a 的第二个rank(列)。

# ix_()函数

可以使用 ix_ 函数来组合不同的向量以获得每个n-uplet的结果。例如,如果要计算从向量a、b和c中的取得的所有三元组的所有a + b * c:

>>> a = np.array([2,3,4,5])
>>> b = np.array([8,5,4])
>>> c = np.array([5,4,6,8,3])
>>> ax,bx,cx = np.ix_(a,b,c)
>>> ax
array([[[2]],
       [[3]],
       [[4]],
       [[5]]])
>>> bx
array([[[8],
        [5],
        [4]]])
>>> cx
array([[[5, 4, 6, 8, 3]]])
>>> ax.shape, bx.shape, cx.shape
((4, 1, 1), (1, 3, 1), (1, 1, 5))
>>> result = ax+bx*cx
>>> result
array([[[42, 34, 50, 66, 26],
        [27, 22, 32, 42, 17],
        [22, 18, 26, 34, 14]],
       [[43, 35, 51, 67, 27],
        [28, 23, 33, 43, 18],
        [23, 19, 27, 35, 15]],
       [[44, 36, 52, 68, 28],
        [29, 24, 34, 44, 19],
        [24, 20, 28, 36, 16]],
       [[45, 37, 53, 69, 29],
        [30, 25, 35, 45, 20],
        [25, 21, 29, 37, 17]]])
>>> result[3,2,4]
17
>>> a[3]+b[2]*c[4]
17

你还可以如下实现reduce:

>>> def ufunc_reduce(ufct, *vectors):
...    vs = np.ix_(*vectors)
...    r = ufct.identity
...    for v in vs:
...        r = ufct(r,v)
...    return r

然后将其用作:

>>> ufunc_reduce(np.add,a,b,c)
array([[[15, 14, 16, 18, 13],
        [12, 11, 13, 15, 10],
        [11, 10, 12, 14,  9]],
       [[16, 15, 17, 19, 14],
        [13, 12, 14, 16, 11],
        [12, 11, 13, 15, 10]],
       [[17, 16, 18, 20, 15],
        [14, 13, 15, 17, 12],
        [13, 12, 14, 16, 11]],
       [[18, 17, 19, 21, 16],
        [15, 14, 16, 18, 13],
        [14, 13, 15, 17, 12]]])

与正常的ufunc.reduce相比,这个版本的reduce的优点是它使用Broadcasting规则,以避免创建参数数组输出的大小乘以向量的数量。

# 使用字符串索引

请参见结构化数组。