$ sudo pip install numpy

$ sudo pip instlal ipython
$ ipython --pylab? #pylab模式下, 會自動導(dǎo)入SciPy, NumPy, Matplotlib模塊

In [1]: a = arange(5)? # 創(chuàng)建數(shù)據(jù)
In [2]: a.dtype
Out[2]: dtype('int64')? # 創(chuàng)建數(shù)組的數(shù)據(jù)類型
In [3]: a.shape? # 數(shù)組的維度, 輸出為tuple
Out[3]: (5,)
In [6]: m = array([[1, 2], [3, 4]])? # array將list轉(zhuǎn)換為NumPy數(shù)組對象
In [7]: m? # 創(chuàng)建多維數(shù)組
Out[7]:
array([[1, 2],
?????? [3, 4]])
In [10]: m.shape? # 維度為2 * 2
Out[10]: (2, 2)
In [14]: m[0, 0]? # 訪問多維數(shù)組中特定位置的元素, 下標(biāo)從0開始
Out[14]: 1
In [15]: m[0, 1]
Out[15]: 2

In [16]: a[2: 4]? # 切片操作類似與Python中l(wèi)ist的切片操作
Out[16]: array([2, 3])
In [18]: a[2 : 5: 2]? # 切片步長為2
Out[18]: array([2, 4])
In [19]: a[ : : -1]? # 翻轉(zhuǎn)數(shù)組
Out[19]: array([4, 3, 2, 1, 0])
In [20]: b = arange(24).reshape(2, 3, 4)? # 修改數(shù)組的維度
In [21]: b.shape
Out[21]: (2, 3, 4)
In [22]: b? # 打印數(shù)組
Out[22]:
array([[[ 0,? 1,? 2,? 3],
??????? [ 4,? 5,? 6,? 7],
??????? [ 8,? 9, 10, 11]],
?????? [[12, 13, 14, 15],
??????? [16, 17, 18, 19],
??????? [20, 21, 22, 23]]])
In [23]: b[1, 2, 3]? # 選取特定元素
Out[23]: 23
In [24]: b[ : , 0, 0]? # 忽略某個(gè)下標(biāo)可以用冒號代替
Out[24]: array([ 0, 12])
In [23]: b[1, 2, 3]
Out[23]: 23
In [24]: b[ : , 0, 0]? # 忽略多個(gè)下標(biāo)可以使用省略號代替
Out[24]: array([ 0, 12])
In [26]: b.ravel()? # 數(shù)組的展平操作
Out[26]:
array([ 0,? 1,? 2,? 3,? 4,? 5,? 6,? 7,? 8,? 9, 10, 11, 12, 13, 14, 15, 16,
?????? 17, 18, 19, 20, 21, 22, 23])
In [27]: b.flatten()? # 與revel功能相同, 這個(gè)函數(shù)會請求分配內(nèi)存來保存結(jié)果
Out[27]:
array([ 0,? 1,? 2,? 3,? 4,? 5,? 6,? 7,? 8,? 9, 10, 11, 12, 13, 14, 15, 16,
?????? 17, 18, 19, 20, 21, 22, 23])
In [30]: b.shape = (6, 4)? # 可以直接對shape屬性賦值元組來設(shè)置維度
In [31]: b
Out[31]:
array([[ 0,? 1,? 2,? 3],
?????? [ 4,? 5,? 6,? 7],
?????? [ 8,? 9, 10, 11],
?????? [12, 13, 14, 15],
?????? [16, 17, 18, 19],
?????? [20, 21, 22, 23]])
In [30]: b.shape = (6, 4)? # 矩陣的轉(zhuǎn)置
In [31]: b
Out[31]:
array([[ 0,? 1,? 2,? 3],
?????? [ 4,? 5,? 6,? 7],
?????? [ 8,? 9, 10, 11],
?????? [12, 13, 14, 15],
?????? [16, 17, 18, 19],
?????? [20, 21, 22, 23]])

In [1]: a = arange(9).reshape(3, 3)? # 生成數(shù)組對象并改變維度
In [2]: a
Out[2]:
array([[0, 1, 2],
?????? [3, 4, 5],
?????? [6, 7, 8]])
In [3]: b = a * 2? # 對a數(shù)組對象所有元素乘2
In [4]: b
Out[4]:
array([[ 0,? 2,? 4],
?????? [ 6,? 8, 10],
?????? [12, 14, 16]])
#######################
In [5]: hstack((a, b))? # 水平組合數(shù)組a和數(shù)組b
Out[5]:
array([[ 0,? 1,? 2,? 0,? 2,? 4],
?????? [ 3,? 4,? 5,? 6,? 8, 10],
?????? [ 6,? 7,? 8, 12, 14, 16]])
??????
In [6]: vstack((a, b))? # 垂直組合數(shù)組a和數(shù)組b
Out[6]:
array([[ 0,? 1,? 2],
?????? [ 3,? 4,? 5],
?????? [ 6,? 7,? 8],
?????? [ 0,? 2,? 4],
?????? [ 6,? 8, 10],
?????? [12, 14, 16]])
In [7]: dstack((a, b))? # 深度組合數(shù)組, 沿z軸方向?qū)盈B組合數(shù)組
Out[7]:
array([[[ 0,? 0],
??????? [ 1,? 2],
??????? [ 2,? 4]],
?????? [[ 3,? 6],
??????? [ 4,? 8],
??????? [ 5, 10]],
?????? [[ 6, 12],
??????? [ 7, 14],
??????? [ 8, 16]]])

In [8]: a
Out[8]:
array([[0, 1, 2],
?????? [3, 4, 5],
?????? [6, 7, 8]])
In [9]: hsplit(a, 3)? # 將數(shù)組沿水平方向分割成三個(gè)相同大小的子數(shù)組
Out[9]:
[array([[0],
??????? [3],
??????? [6]]),
?array([[1],
??????? [4],
??????? [7]]),
?array([[2],
??????? [5],
??????? [8]])]
In [10]: vsplit(a, 3)? # 將數(shù)組沿垂直方向分割成三個(gè)子數(shù)組
Out[10]: [array([[0, 1, 2]]), array([[3, 4, 5]]), array([[6, 7, 8]])]

In [12]: a.ndim? # 給出數(shù)組的尾數(shù)或數(shù)組的軸數(shù)
Out[12]: 2
In [13]: a.size? # 數(shù)組中元素的個(gè)數(shù)
Out[13]: 9
In [14]: a.itemsize? # 數(shù)組中元素在內(nèi)存中所占字節(jié)數(shù)(int64)
Out[14]: 8
In [15]: a.nbytes? # 數(shù)組所占總字節(jié)數(shù), size * itemsize
Out[15]: 72
In [18]: a.T? # 和transpose函數(shù)一樣, 求數(shù)組的轉(zhuǎn)置
Out[18]:
array([[0, 3, 6],
?????? [1, 4, 7],
?????? [2, 5, 8]])

In [19]: a.tolist()? # 將NumPy數(shù)組轉(zhuǎn)換成python中的list
Out[19]: [[0, 1, 2], [3, 4, 5], [6, 7, 8]]

In [22]: c = eye(2)? # 構(gòu)建2維單位矩陣
In [23]: c
Out[23]:
array([[ 1.,? 0.],
?????? [ 0.,? 1.]])
In [25]: savetxt("eye.txt", c)? # 將矩陣保存到文件中
In [5]: c, v = loadtxt("test.csv", delimiter=",", usecols=(0, 1), unpack=True)? # 分隔符為, usecols為元組表示要獲取的字段數(shù)據(jù)(每一行的第零段和第一段), unpack為True表示拆分存儲不同列的數(shù)據(jù), 分別存入c, v
In [12]: c
Out[12]: array([ 1.,? 4.,? 7.])
In [13]: mean(c)? # 計(jì)算矩陣c的mean均值
Out[13]: 4.0
In [14]: np.max(c)? # 求數(shù)組中的最大值
Out[14]: 7.0
In [15]: np.min(c)? # 求數(shù)組中的最小值
Out[15]: 1.0
In [16]: np.ptp(c)? # 返回?cái)?shù)組最大值和最小值之間的差值
Out[16]: 6.0
In [18]: numpy.median(c)? # 找到數(shù)組中的中位數(shù)(中間兩個(gè)數(shù)的平均值)
Out[18]: 4.0
In [19]: numpy.var(c)? # 計(jì)算數(shù)組的方差
Out[19]: 6.0
In [20]: numpy.diff(c)? # 返回相鄰數(shù)組元素的差值構(gòu)成的數(shù)組
Out[20]: array([ 3.,? 3.])
In [21]: numpy.std(c)? # 計(jì)算數(shù)組的標(biāo)準(zhǔn)差
Out[21]: 2.4494897427831779
In [22]: numpy.where(c > 3)? # 返回滿足條件的數(shù)組元素的下標(biāo)組成的數(shù)組
Out[22]: (array([1, 2]),)