这些有用的片段在面试中会经常出现，也能够作为日常的numpy练习。
1、导入numpy
import numpy as np2、打印numpy信息
print(np.__version__)np.show_config()3、创立空向量
Z = np.zeros(10)print(Z)4、获取numpy 函数的文档
python -c "import numpy; numpy.info(numpy.add)"5、创立大小为10但第5个值为1的空向量
Z = np.zeros(10)Z[4] = 1print(Z)6、创立一个值从10到49的向量
Z = np.arange(10,50)print(Z)7、反转一个向量(第一个元素变成最初一个元素)
Z = np.arange(50)Z = Z[::-1]8、创立一个值从0到8的3x3矩阵
Z = np.arange(9).reshape(3,3)print(Z)9、从[1,2,0,0,4,0]中找到非零元素的下标
nz = np.nonzero([1,2,0,0,4,0])print(nz)10、创立一个3x3单位矩阵
Z = np.eye(3)print(Z)11、创立一个带有随机值的3x3x3数组
Z = np.random.random((3,3,3))print(Z)12、创立一个带有随机值的10x10数组，并找到最小值和最大值
Z = np.random.random((10,10))Zmin, Zmax = Z.min(), Z.max()print(Zmin, Zmax)13、创立一个大小为30的随机向量，并找出平均值
Z = np.random.random(30)m = Z.mean()print(m)14、创立一个边界为1，外部为0的2d数组
Z = np.ones((10,10))Z[1:-1,1:-1] = 015 、上面表达式的后果是什么?
0 * np.nannp.nan == np.nan np.inf > np.nannp.nan - np.nan0.3 == 3 * 0.116、创立一个5 × 5矩阵，对角线值为1,2,3,4
Z = np.diag(1+np.arange(4),k=-1)print(Z)17、创立一个8x8的矩阵，并应用0，1距离填充
Z = np.zeros((8,8),dtype=int)Z[1::2,::2] = 1Z[::2,1::2] = 1print(Z)18、一个(6,7,8)形态数组，第100个元素的下标(x,y,z)是多少?
print(np.unravel_index(100,(6,7,8)))19、应用tile函数创立一个8x8的0，1距离填充矩阵
Z = np.tile( np.array([[0,1],[1,0]]), (4,4))print(Z)20、标准化一个5x5随机矩阵
Z = np.random.random((5,5))Zmax, Zmin = Z.max(), Z.min()Z = (Z - Zmin)/(Zmax - Zmin)print(Z)21、创立一个自定义的dtype，将色彩形容为4个unisgned字节(RGBA)
color = np.dtype([("r", np.ubyte, 1),("g", np.ubyte, 1),("b", np.ubyte, 1),("a", np.ubyte, 1)])22、将一个5x3矩阵乘以一个3x2矩阵(实矩阵乘积)
Z = np.dot(np.ones((5,3)), np.ones((3,2)))print(Z)23、给定一个一维数组，对3到8之间的所有元素求反、
# Author: Evgeni BurovskiZ = np.arange(11)Z[(3 < Z) & (Z <= 8)] *= -124、以下脚本的输入是什么?
# Author: Jake VanderPlasprint(sum(range(5),-1))from numpy import *print(sum(range(5),-1))##后果 ：1025、整数向量Z，上面哪个表达式是非法的?
Z**Z2 << Z >> 2Z <- Z1j*ZZ/1/1Z<Z>Z #除这个外都非法26、下列表达式的后果是什么?
np.array(0) // np.array(0)np.array(0) // np.array(0.)np.array(0) / np.array(0)np.array(0) / np.array(0.)27、如何四舍五入?
# Author: Charles R HarrisZ = np.random.uniform(-10,+10,10)print (np.trunc(Z + np.copysign(0.5, Z)))28、 应用 5 种不同的办法提取随机数组的整数局部
Z = np.random.uniform(0,10,10)print (Z - Z%1)print (np.floor(Z))print (np.ceil(Z)-1)print (Z.astype(int))print (np.trunc(Z))29、 创立一个值范畴为 0 到 4的 5x5 矩阵
Z = np.zeros((5,5))Z += np.arange(5)print(Z)30 、 创立生成器函数，生成 10 个整数并应用它来构建一个数组
def generate():for x in xrange(10):yield xZ = np.fromiter(generate(),dtype=float,count=-1)print(Z)31、 创立一个大小为 10 的向量，其值范畴为 0 到 1（均不包含在内）
Z = np.linspace(0,1,12,endpoint=True)[1:-1]print(Z)32、 创立一个大小为 10 的随机向量并对其进行排序
Z = np.random.random(10)Z.sort()print(Z)33、 如何比 np.sum 更快地求和一个小数组？
# Author: Evgeni BurovskiZ = np.arange(10)np.add.reduce(Z)34、 思考查看两个随机数组 A 和 B是否相等
A = np.random.randint(0,2,5)B = np.random.randint(0,2,5)equal = np.allclose(A,B)print(equal)35、 使数组不可变（只读）
Z = np.zeros(10)Z.flags.writeable = FalseZ[0] = 136、 一个示意笛卡尔坐标的随机 10x2 矩阵，将它们转换为极坐标
X,Y = Z[:,0], Z[:,1]R = np.sqrt(X**2+Y**2)T = np.arctan2(Y,X)print(R)print(T)37、 创立大小为 10 的随机向量并将最大值替换为 0
Z = np.random.random(10)Z[Z.argmax()] = 0print(Z)38、 创立一个结构化数组，其中 x 和坐标笼罩 [0,1]x[0,1] 区域
Z = np.zeros((10,10), [('x',float),('y',float)])Z['x'], Z['y'] = np.meshgrid(np.linspace(0,1,10),np.linspace(0,1,10))print(Z)39、 给定两个数组 X 和 Y，结构柯西矩阵 C (Cij = 1/(xi yj))
# Author: Evgeni BurovskiX = np.arange(8)Y = X + 0.5C = 1.0 / np.subtract.outer(X, Y)print(np.linalg.det(C))40 、 打印每个 numpy 标量类型的最小和最大可示意值
for dtype in [np.int8, np.int32, np.int64]:print(np.iinfo(dtype).min)print(np.iinfo(dtype).max)for dtype in [np.float32, np.float64]:print(np.finfo(dtype).min)print(np.finfo(dtype).max)print(np.finfo(dtype).eps)41、 如何打印数组的所有值？
np.set_printoptions(threshold=np.nan)Z = np.zeros((25,25))print(Z)42、 如何在数组中找到最靠近的值（到给定的标量）？
Z = np.arange(100)v = np.random.uniform(0,100)index = (np.abs(Z-v)).argmin()print(Z[index])43 、 创立示意地位 (x,y) 和色彩 (r,g,b) 的结构化数组
Z = np.zeros(10, [ ('position', [ ('x', float, 1),('y', float, 1)]),('color', [ ('r', float, 1),('g', float, 1),('b', float, 1)])])print(Z)44、 逐点查找形态为 (100,2) 的随机向量示意坐标的间隔
Z = np.random.random((10,2))X,Y = np.atleast_2d(Z[:,0]), np.atleast_2d(Z[:,1])D = np.sqrt( (X-X.T)**2 + (Y-Y.T)**2)print(D)# Much faster with scipyimport scipy# Thanks Gavin Heverly-Coulson (#issue 1)import scipy.spatialZ = np.random.random((10,2))D = scipy.spatial.distance.cdist(Z,Z)print(D)45 、 如何将浮点（32位）数组转换为整数（32位）？
Z = np.arange(10, dtype=np.int32)Z = Z.astype(np.float32, copy=False)46 、 读取上面格局的文件？
# -------------1,2,3,4,56,,,7,8,,9,10,11# -------------Z = np.genfromtxt("missing.dat", delimiter=",")47 、 numpy 数组的枚举能够用什么其余模式示意？
Z = np.arange(9).reshape(3,3)for index, value in np.ndenumerate(Z):print(index, value)for index in np.ndindex(Z.shape):48、生成一个二维类高斯数组
X, Y = np.meshgrid(np.linspace(-1,1,10), np.linspace(-1,1,10))D = np.sqrt(X*X+Y*Y)sigma, mu = 1.0, 0.0G = np.exp(-( (D-mu)**2 / ( 2.0 * sigma**2 ) ) )print(G)49 、 如何在二维数组中随机搁置 p 个元素？
# Author: Divakarn = 10p = 3Z = np.zeros((n,n))np.put(Z, np.random.choice(range(n*n), p, replace=False),1)50、矩阵减去每一行的均值
# Author: Warren WeckesserX = np.random.rand(5, 10)# Recent versions of numpyY = X - X.mean(axis=1, keepdims=True)# Older versions of numpyY = X - X.mean(axis=1).reshape(-1, 1)51、如何排序数组的第n列?
# Author: Steve TjoaZ = np.random.randint(0,10,(3,3))print(Z)print(Z[Z[:,1].argsort()])52、如何判断一个给定的2D数组有空列?
# Author: Warren WeckesserZ = np.random.randint(0,3,(3,10))print((~Z.any(axis=0)).any())53、在数组中找到与给定值最近的值
Z = np.random.uniform(0,1,10)z = 0.5m = Z.flat[np.abs(Z - z).argmin()]print(m)54、创立一个具备name属性的数组类
class NamedArray(np.ndarray):def __new__(cls, array, name="no name"):obj = np.asarray(array).view(cls)obj.name = namereturn objdef __array_finalize__(self, obj):if obj is None: returnself.info = getattr(obj, 'name', "no name")Z = NamedArray(np.arange(10), "range_10")print (Z.name)55、一个给定的向量，如何给第二个向量索引的每个元素加1(蕴含反复索引)?
# Author: Brett OlsenZ = np.ones(10)I = np.random.randint(0,len(Z),20)Z += np.bincount(I, minlength=len(Z))print(Z)56、如何积加一个向量(X)的元素到一个数组(F)基于索引列表(I)?
# Author: Alan G IsaacX = [1,2,3,4,5,6]I = [1,3,9,3,4,1]F = np.bincount(I,X)print(F)57、一个(w,h,3)图像(dtype=ubyte)，计算惟一色彩的数量
# Author: Nadav Horeshw,h = 16,16I = np.random.randint(0,2,(h,w,3)).astype(np.ubyte)F = I[...,0]*256*256 + I[...,1]*256 +I[...,2]n = len(np.unique(F))print(np.unique(I))58、思考一个四维数组，如何同时失去最初两个轴的和?
A = np.random.randint(0,10,(3,4,3,4))sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)print(sum)59、一个一维向量D，应用子集雷同大小的向量S来计算D子集的平均值?
# Author: Jaime Fernández del RíoD = np.random.uniform(0,1,100)S = np.random.randint(0,10,100)D_sums = np.bincount(S, weights=D)D_counts = np.bincount(S)D_means = D_sums / D_countsprint(D_means)60 、 如何取得点积的对角线？
# Author: Mathieu Blondel# Slow versionnp.diag(np.dot(A, B))# Fast versionnp.sum(A * B.T, axis=1)# Faster versionnp.einsum("ij,ji->i", A, B).61、 向量 [1, 2, 3, 4, 5]，构建一个新向量，每个值之间有 3 个间断的0？
# Author: Warren WeckesserZ = np.array([1,2,3,4,5])nz = 3Z0 = np.zeros(len(Z) + (len(Z)-1)*(nz))Z0[::nz+1] = Zprint(Z0)62、 维度为 (5,5,3) 的数组，将它乘以一个维度为 (5,5) 的数组？
A = np.ones((5,5,3))B = 2*np.ones((5,5))print(A * B[:,:,None])63、 替换数组的两行？
# Author: Eelco HoogendoornA = np.arange(25).reshape(5,5)A[[0,1]] = A[[1,0]]print(A)64、 一组形容 10 个三角形（具备共享顶点）的 10 个三元组，找到组成所有三角形的惟一线段集
# Author: Nicolas P. Rougierfaces = np.random.randint(0,100,(10,3))F = np.roll(faces.repeat(2,axis=1),-1,axis=1)F = F.reshape(len(F)*3,2)F = np.sort(F,axis=1)G = F.view( dtype=[('p0',F.dtype),('p1',F.dtype)] )G = np.unique(G)print(G)65 、 一个 bincount数组 C，如何生成一个数组 A 使得np.bincount(A) == C?
# Author: Jaime Fernández del RíoC = np.bincount([1,1,2,3,4,4,6])A = np.repeat(np.arange(len(C)), C)print(A)66、 应用数组上的滑动窗口计算平均值？
# Author: Jaime Fernández del Ríodef moving_average(a, n=3) :ret = np.cumsum(a, dtype=float)ret[n:] = ret[n:] - ret[:-n]return ret[n - 1:] / nZ = np.arange(20)print(moving_average(Z, n=3))67、 一维数组 Z，构建一个二维数组，其第一行是 (Z[0],Z[1],Z[2])，随后的每一行挪动 1（最初一行应该是 (Z[3] ,Z[2],Z[1])
# Author: Joe Kington / Erik Rigtorpfrom numpy.lib import stride_tricksdef rolling(a, window):shape = (a.size - window + 1, window)strides = (a.itemsize, a.itemsize)return stride_tricks.as_strided(a, shape=shape, strides=strides)Z = rolling(np.arange(10), 3)print(Z)68、 如何对布尔值取反，或更改浮点数的符号？
# Author: Nathaniel J. SmithZ = np.random.randint(0,2,100)np.logical_not(arr, out=arr)Z = np.random.uniform(-1.0,1.0,100)np.negative(arr, out=arr)69 、 思考 2 组点 P0,P1 形容线 (2d) 和一个点 p，如何计算从 p 到每条线 i (P0[i],P1[i]) 的间隔？
def distance(P0, P1, p):T = P1 - P0L = (T**2).sum(axis=1)U = -((P0[:,0]-p[...,0])*T[:,0] + (P0[:,1]-p[...,1])*T[:,1]) / LU = U.reshape(len(U),1)D = P0 + U*T - preturn np.sqrt((D**2).sum(axis=1))P0 = np.random.uniform(-10,10,(10,2))P1 = np.random.uniform(-10,10,(10,2))p = np.random.uniform(-10,10,( 1,2))print(distance(P0, P1, p))70 、 2 组点 P0,P1 形容线 (2d) 和一组点 P，如何计算从每个点 j (P[j]) 到每条线 i (P0[i],P1[i]) 的间隔？
# Author: Italmassov Kuanysh# based on distance function from previous questionP0 = np.random.uniform(-10, 10, (10,2))P1 = np.random.uniform(-10,10,(10,2))p = np.random.uniform(-10, 10, (10,2))print np.array([distance(P0,P1,p_i) for p_i in p])71、 一个任意数组，编写一个函数来提取具备固定形态并以给定元素为核心的子局部（必要时能够应用填充值填充）
# Author: Nicolas RougierZ = np.random.randint(0,10,(10,10))shape = (5,5)fill = 0position = (1,1)R = np.ones(shape, dtype=Z.dtype)*fillP = np.array(list(position)).astype(int)Rs = np.array(list(R.shape)).astype(int)Zs = np.array(list(Z.shape)).astype(int)R_start = np.zeros((len(shape),)).astype(int)R_stop = np.array(list(shape)).astype(int)Z_start = (P-Rs//2)Z_stop = (P+Rs//2)+Rs%2R_start = (R_start - np.minimum(Z_start,0)).tolist()Z_start = (np.maximum(Z_start,0)).tolist()R_stop = np.maximum(R_start, (R_stop - np.maximum(Z_stop-Zs,0))).tolist()Z_stop = (np.minimum(Z_stop,Zs)).tolist()r = [slice(start,stop) for start,stop in zip(R_start,R_stop)]z = [slice(start,stop) for start,stop in zip(Z_start,Z_stop)]R[r] = Z[z]print(Z)print(R)72、 一个数组 Z = [1,2,3,4,5,6,7,8,9,10,11,12,13,14]，如何生成一个数组 R = [[1,2,3, 4], [2,3,4,5], [3,4,5,6], ..., [11,12,13,14]]？
# Author: Stefan van der WaltZ = np.arange(1,15,dtype=uint32)R = stride_tricks.as_strided(Z,(11,4),(4,4))print(R)73、 计算矩阵秩
# Author: Stefan van der WaltZ = np.random.uniform(0,1,(10,10))U, S, V = np.linalg.svd(Z) # Singular Value Decompositionrank = np.sum(S > 1e-10)74、 如何找到数组中呈现频率最高的值？
Z = np.random.randint(0,10,50)print(np.bincount(Z).argmax())75 、 从随机 10x10 矩阵中提取所有间断的 3x3 块
# Author: Chris BarkerZ = np.random.randint(0,5,(10,10))n = 3i = 1 + (Z.shape[0]-3)j = 1 + (Z.shape[1]-3)C = stride_tricks.as_strided(Z, shape=(i, j, n, n), strides=Z.strides + Z.strides)print(C)76 、 创立一个二维数组子类，使得 Z[i,j] == Z[j,i]
# Author: Eric O. Lebigot# Note: only works for 2d array and value setting using indicesclass Symetric(np.ndarray):def __setitem__(self, (i,j), value):super(Symetric, self).__setitem__((i,j), value)super(Symetric, self).__setitem__((j,i), value)def symetric(Z):return np.asarray(Z + Z.T - np.diag(Z.diagonal())).view(Symetric)S = symetric(np.random.randint(0,10,(5,5)))S[2,3] = 42print(S)77、 一组形态为 (n,n) 的 p 个矩阵和一组形态为 (n,1) 的 p 个向量、 如何一次计算 p 个矩阵乘积之和？（后果具备形态（n，1））
# Author: Stefan van der Waltp, n = 10, 20M = np.ones((p,n,n))V = np.ones((p,n,1))S = np.tensordot(M, V, axes=[[0, 2], [0, 1]])print(S)# It works, because:# M is (p,n,n)# V is (p,n,1)# Thus, summing over the paired axes 0 and 0 (of M and V independently),# and 2 and 1, to remain with a (n,1) vector.78 、 思考一个 16x16 数组，如何取得块的和（块大小为 4x4）？
# Author: Robert KernZ = np.ones(16,16)k = 4S = np.add.reduceat(np.add.reduceat(Z, np.arange(0, Z.shape[0], k), axis=0),np.arange(0, Z.shape[1], k), axis=1)79 、 如何应用 numpy 数组实现 Game of Life？
# Author: Nicolas Rougierdef iterate(Z):# Count neighboursN = (Z[0:-2,0:-2] + Z[0:-2,1:-1] + Z[0:-2,2:] +Z[1:-1,0:-2] + Z[1:-1,2:] +Z[2: ,0:-2] + Z[2: ,1:-1] + Z[2: ,2:])# Apply rulesbirth = (N==3) & (Z[1:-1,1:-1]==0)survive = ((N==2) | (N==3)) & (Z[1:-1,1:-1]==1)Z[...] = 0Z[1:-1,1:-1][birth | survive] = 1return ZZ = np.random.randint(0,2,(50,50))for i in range(100): Z = iterate(Z)80、取得数组的n个最大的值
Z = np.arange(10000)np.random.shuffle(Z)n = 5# Slowprint (Z[np.argsort(Z)[-n:]])# Fastprint (Z[np.argpartition(-Z,n)[:n]])81、构建笛卡尔积(每个项的每个组合)
# Author: Stefan Van der Waltdef cartesian(arrays):arrays = [np.asarray(a) for a in arrays]shape = (len(x) for x in arrays)ix = np.indices(shape, dtype=int)ix = ix.reshape(len(arrays), -1).Tfor n, arr in enumerate(arrays):ix[:, n] = arrays[n][ix[:, n]]return ixprint (cartesian(([1, 2, 3], [4, 5], [6, 7])))82、从一个惯例数组创立一个records数组?
Z = np.array([("Hello", 2.5, 3),("World", 3.6, 2)])R = np.core.records.fromarrays(Z.T,names='col1, col2, col3',formats = 'S8, f8, i8')83、一个大向量Z，应用3种不同的办法计算Z的3次方
Author: Ryan G.x = np.random.rand(5e7)%timeit np.power(x,3)1 loops, best of 3: 574 ms per loop%timeit x*x*x1 loops, best of 3: 429 ms per loop%timeit np.einsum('i,i,i->i',x,x,x)1 loops, best of 3: 244 ms per loop84、形态为(8,3)和(2,2)的两个数组A和B、如何在A中找到蕴含B每一行元素的行不论B中元素的程序是什么?
# Author: Gabe SchwartzA = np.random.randint(0,5,(8,3))B = np.random.randint(0,5,(2,2))C = (A[..., np.newaxis, np.newaxis] == B)rows = (C.sum(axis=(1,2,3)) >= B.shape[1]).nonzero()[0]print(rows)85、一个10x3矩阵，提取不相等值的行(例如[2,2,3])
# Author: Robert KernZ = np.random.randint(0,5,(10,3))E = np.logical_and.reduce(Z[:,1:] == Z[:,:-1], axis=1)U = Z[~E]print(Z)print(U)86、将一个int型向量转换为一个矩阵二进制示意模式
# Author: Warren WeckesserI = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128])B = ((I.reshape(-1,1) & (2**np.arange(8))) != 0).astype(int)print(B[:,::-1])# Author: Daniel T. McDonaldI = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128], dtype=np.uint8)print(np.unpackbits(I[:, np.newaxis], axis=1))87、给定一个二维数组，如何提取惟一的行?
# Author: Jaime Fernández del RíoZ = np.random.randint(0,2,(6,3))T = np.ascontiguousarray(Z).view(np.dtype((np.void, Z.dtype.itemsize * Z.shape[1])))_, idx = np.unique(T, return_index=True)uZ = Z[idx]print(uZ)88、思考两个向量A和B，应用einsum求sum、* 、inner、outer
# Author: Alex Riley# Make sure to read: http://ajcr.net/Basic-guide-to-einsum/np.einsum('i->', A) # np.sum(A)np.einsum('i,i->i', A, B) # A * Bnp.einsum('i,i', A, B) # np.inner(A, B)np.einsum('i,j', A, B) # np.outer(A, B)89 、 由两个向量（X，Y）形容的门路，如何应用等距样本对其进行采样？
# Author: Bas Swinckelsphi = np.arange(0, 10*np.pi, 0.1)a = 1x = a*phi*np.cos(phi)y = a*phi*np.sin(phi)dr = (np.diff(x)**2 + np.diff(y)**2)**.5 # segment lengthsr = np.zeros_like(x)r[1:] = np.cumsum(dr) # integrate pathr_int = np.linspace(0, r.max(), 200) # regular spaced pathx_int = np.interp(r_int, r, x) # integrate pathy_int = np.interp(r_int, r, y)90 、 给定一个整数 n 和一个二维数组 X，从 X 中抉择能够解释为从具备 n 度的多项散布中抽取的行，即仅蕴含整数且总和为 n 的行、
# Author: Evgeni BurovskiX = np.asarray([[1.0, 0.0, 3.0, 8.0],[2.0, 0.0, 1.0, 1.0],[1.5, 2.5, 1.0, 0.0]])n = 4M = np.logical_and.reduce(np.mod(X, 1) == 0, axis=-1)M &= (X.sum(axis=-1) == n)print(X[M])以上代码整顿自：https://github.com/rougier/nu... 有删减和批改，需要的话能够star原我的项目