>>> a = np.ones((2,3), dtype=int)
>>> b = np.random.random((2,3))
>>> a *= 3
>>> a
array([[3, 3, 3],
[3, 3, 3]])
>>> b += a
>>> b
array([[ 3.417022 , 3.72032449, 3.00011437],
[ 3.30233257, 3.14675589, 3.09233859]])
>>> a += b # b is not automatically converted to integer type
Traceback (most recent call last):
...
TypeError: Cannot cast ufunc add output from dtype('float64') to dtype('int64') with casting rule 'same_kind'
>>> a = np.arange(10)**3
>>> a
array([ 0, 1, 8, 27, 64, 125, 216, 343, 512, 729])
>>> a[2]
8
>>> a[2:5]
array([ 8, 27, 64])
>>> a[:6:2] = -1000 # equivalent to a[0:6:2] = -1000; from start to position 6, exclusive, set every 2nd element to -1000
>>> a
array([-1000, 1, -1000, 27, -1000, 125, 216, 343, 512, 729])
>>> a[ : :-1] # reversed a
array([ 729, 512, 343, 216, 125, -1000, 27, -1000, 1, -1000])
>>> for i in a:
... print(i**(1/3.))
...
nan
1.0
nan
3.0
nan
5.0
6.0
7.0
8.0
9.0
多维数组可以每个轴有一个索引。这些索引由一个逗号分割的元组给出。
>>> def f(x,y):
... return 10*x+y
...
>>> b = fromfunction(f,(5,4),dtype=int)
>>> b
array([[ 0, 1, 2, 3],
[10, 11, 12, 13],
[20, 21, 22, 23],
[30, 31, 32, 33],
[40, 41, 42, 43]])
>>> b[2,3]
23
>>> b[0:5, 1] # each row in the second column of b
array([ 1, 11, 21, 31, 41])
>>> b[ : ,1] # equivalent to the previous example
array([ 1, 11, 21, 31, 41])
>>> b[1:3, : ] # each column in the second and third row of b
array([[10, 11, 12, 13],
[20, 21, 22, 23]])
当少于轴数的索引被提供时，确失的索引被认为是整个切片：
>>> b[-1] # the last row. Equivalent to b[-1,:]
array([40, 41, 42, 43])
>>> a = arange(12)
>>> b = a # no new object is created
>>> b is a # a and b are two names for the same ndarray object
True
>>> b.shape = 3,4 # changes the shape of a
>>> a.shape
(3, 4)
Python 传递不定对象作为参考(形参)，所以函数调用不拷贝数组。
>>> def f(x):
... print id(x)
...
>>> id(a) # id is a unique identifier of an object
148293216
>>> f(a)
148293216
视图(view)和浅复制
不同的数组对象分享同一个数据。视图方法创造一个新的数组对象指向同一数据。
>>> c = a.view()
>>> c is a
False
>>> c.base is a # c is a view of the data owned by a
True
>>> c.flags.owndata
False
>>>
>>> c.shape = 2,6 # a's shape doesn't change
>>> a.shape
(3, 4)
>>> c[0,4] = 1234 # a's data changes
>>> a
array([[ 0, 1, 2, 3],
[1234, 5, 6, 7],
[ 8, 9, 10, 11]])
切片数组返回它的一个视图：
>>> s = a[ : , 1:3] # spaces added for clarity; could also be written "s = a[:,1:3]"
>>> s[:] = 10 # s[:] is a view of s. Note the difference between s=10 and s[:]=10
>>> a
array([[ 0, 10, 10, 3],
[1234, 10, 10, 7],
[ 8, 10, 10, 11]])
深复制
这个复制方法完全复制数组和它的数据。
>>> d = a.copy() # a new array object with new data is created
>>> d is a
False
>>> d.base is a # d doesn't share anything with a
False
>>> d[0,0] = 9999
>>> a
array([[ 0, 10, 10, 3],
[1234, 10, 10, 7],
[ 8, 10, 10, 11]])
>>> a = arange(12)**2 # the first 12 square numbers
>>> i = 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 = array( [ [ 3, 4], [ 9, 7 ] ] ) # a bidimensional array of indices
>>> a[j] # the same shape as j
array([[ 9, 16],
[81, 49]])
>>> palette = array( [ [0,0,0], # black
... [255,0,0], # red
... [0,255,0], # green
... [0,0,255], # blue
... [255,255,255] ] ) # white
>>> image = 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]]])
我们也可以给出不不止一维的索引，每一维的索引数组必须有相同的形状。
>>> a = arange(12).reshape(3,4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> i = array( [ [0,1], # indices for the first dim of a
... [1,2] ] )
>>> j = 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]]])
自然，我们可以把i和j放到序列中(比如说列表)然后通过list索引。
>>> l = [i,j]
>>> a[l] # equivalent to a[i,j]
array([[ 2, 5],
[ 7, 11]])
然而，我们不能把i和j放在一个数组中，因为这个数组将被解释成索引a的第一维。
>>> s = array( [i,j] )
>>> a[s] # not what we want
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-100-b912f631cc75> in <module>()
----> 1 a[s]
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]])
另一个常用的数组索引用法是搜索时间序列最大值。
>>> time = linspace(20, 145, 5) # time scale
>>> data = sin(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, xrange(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 ])
>>>
>>> all(data_max == data.max(axis=0))
True
你也可以使用数组索引作为目标来赋值：
>>> a = arange(5)
>>> a
array([0, 1, 2, 3, 4])
>>> a[[1,3,4]] = 0
>>> a
array([0, 0, 2, 0, 0])
然而，当一个索引列表包含重复时，赋值被多次完成，保留最后的值：
>>> a = arange(5)
>>> a[[0,0,2]]=[1,2,3]
>>> a
array([2, 1, 3, 3, 4])
这足够合理，但是小心如果你想用Python的+=结构，可能结果并非你所期望：
>>> a = arange(5)
>>> a[[0,0,2]]+=1
>>> a
array([1, 1, 3, 3, 4])
>>> a = 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]], dtype=bool)
>>> a[b] # 1d array with the selected elements
array([ 5, 6, 7, 8, 9, 10, 11])
这个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]])
import numpy
import pylab
# Build a vector of 10000 normal deviates with variance 0.5^2 and mean 2
mu, sigma = 2, 0.5
v = numpy.random.normal(mu,sigma,10000)
# Plot a normalized histogram with 50 bins
pylab.hist(v, bins=50, normed=1) # matplotlib version (plot)
pylab.show()
# Compute the histogram with numpy and then plot it
(n, bins) = numpy.histogram(v, bins=50, normed=True) # NumPy version (no plot)
pylab.plot(.5*(bins[1:]+bins[:-1]), n)
pylab.show()
