美文网首页
numpy 100道通关题(三)

numpy 100道通关题(三)

作者: AlexDM | 来源:发表于2020-01-03 10:54 被阅读0次
    NumPy

    64. Consider a given vector, how to add 1 to each element indexed by a second vector (be careful with repeated indices)? (★★★)

    # Author: Brett Olsen
    
    Z = np.ones(10)
    I = np.random.randint(0,len(Z),20)
    Z += np.bincount(I, minlength=len(Z))
    print(Z)
    
    # Another solution
    # Author: Bartosz Telenczuk
    np.add.at(Z, I, 1)
    print(Z)
    

    65. How to accumulate elements of a vector (X) to an array (F) based on an index list (I)? (★★★)

    # Author: Alan G Isaac
    
    X = [1,2,3,4,5,6]
    I = [1,3,9,3,4,1]
    F = np.bincount(I,X)
    print(F)
    

    66. Considering a (w,h,3) image of (dtype=ubyte), compute the number of unique colors (★★★)

    # Author: Nadav Horesh
    
    w,h = 16,16
    I = 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))
    

    67. Considering a four dimensions array, how to get sum over the last two axis at once? (★★★)

    A = np.random.randint(0,10,(3,4,3,4))
    # solution by passing a tuple of axes (introduced in numpy 1.7.0)
    sum = A.sum(axis=(-2,-1))
    print(sum)
    # solution by flattening the last two dimensions into one
    # (useful for functions that don't accept tuples for axis argument)
    sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)
    print(sum)
    

    68. Considering a one-dimensional vector D, how to compute means of subsets of D using a vector S of same size describing subset indices? (★★★)

    # Author: Jaime Fernández del Río
    
    D = 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_counts
    print(D_means)
    
    # Pandas solution as a reference due to more intuitive code
    import pandas as pd
    print(pd.Series(D).groupby(S).mean())
    

    69. How to get the diagonal of a dot product? (★★★)

    # Author: Mathieu Blondel
    
    A = np.random.uniform(0,1,(5,5))
    B = np.random.uniform(0,1,(5,5))
    
    # Slow version  
    np.diag(np.dot(A, B))
    
    # Fast version
    np.sum(A * B.T, axis=1)
    
    # Faster version
    np.einsum("ij,ji->i", A, B)
    

    70. Consider the vector [1, 2, 3, 4, 5], how to build a new vector with 3 consecutive zeros interleaved between each value? (★★★)

    # Author: Warren Weckesser
    
    Z = np.array([1,2,3,4,5])
    nz = 3
    Z0 = np.zeros(len(Z) + (len(Z)-1)*(nz))
    Z0[::nz+1] = Z
    print(Z0)
    

    71. Consider an array of dimension (5,5,3), how to mulitply it by an array with dimensions (5,5)? (★★★)

    A = np.ones((5,5,3))
    B = 2*np.ones((5,5))
    print(A * B[:,:,None])
    

    72. How to swap two rows of an array? (★★★)

    # Author: Eelco Hoogendoorn
    
    A = np.arange(25).reshape(5,5)
    A[[0,1]] = A[[1,0]]
    print(A)
    

    73. Consider a set of 10 triplets describing 10 triangles (with shared vertices), find the set of unique line segments composing all the triangles (★★★)

    # Author: Nicolas P. Rougier
    
    faces = 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)
    

    74. Given an array C that is a bincount, how to produce an array A such that np.bincount(A) == C? (★★★)

    # Author: Jaime Fernández del Río
    
    C = np.bincount([1,1,2,3,4,4,6])
    A = np.repeat(np.arange(len(C)), C)
    print(A)
    

    75. How to compute averages using a sliding window over an array? (★★★)

    # Author: Jaime Fernández del Río
    
    def moving_average(a, n=3) :
        ret = np.cumsum(a, dtype=float)
        ret[n:] = ret[n:] - ret[:-n]
        return ret[n - 1:] / n
    Z = np.arange(20)
    print(moving_average(Z, n=3))
    

    76. Consider a one-dimensional array Z, build a two-dimensional array whose first row is (Z[0],Z[1],Z[2]) and each subsequent row is shifted by 1 (last row should be (Z[-3],Z[-2],Z[-1]) (★★★)

    # Author: Joe Kington / Erik Rigtorp
    from numpy.lib import stride_tricks
    
    def 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)
    

    77. How to negate a boolean, or to change the sign of a float inplace? (★★★)

    # Author: Nathaniel J. Smith
    
    Z = np.random.randint(0,2,100)
    np.logical_not(Z, out=Z)
    
    Z = np.random.uniform(-1.0,1.0,100)
    np.negative(Z, out=Z)
    

    78. Consider 2 sets of points P0,P1 describing lines (2d) and a point p, how to compute distance from p to each line i (P0[i],P1[i])? (★★★)

    def distance(P0, P1, p):
        T = P1 - P0
        L = (T**2).sum(axis=1)
        U = -((P0[:,0]-p[...,0])*T[:,0] + (P0[:,1]-p[...,1])*T[:,1]) / L
        U = U.reshape(len(U),1)
        D = P0 + U*T - p
        return 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))
    

    79. Consider 2 sets of points P0,P1 describing lines (2d) and a set of points P, how to compute distance from each point j (P[j]) to each line i (P0[i],P1[i])? (★★★)

    # Author: Italmassov Kuanysh
    
    # based on distance function from previous question
    P0 = 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]))
    

    80. Consider an arbitrary array, write a function that extract a subpart with a fixed shape and centered on a given element (pad with a fill value when necessary) (★★★)

    # Author: Nicolas Rougier
    
    Z = np.random.randint(0,10,(10,10))
    shape = (5,5)
    fill  = 0
    position = (1,1)
    
    R = np.ones(shape, dtype=Z.dtype)*fill
    P  = 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%2
    
    R_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)
    

    81. Consider an array Z = [1,2,3,4,5,6,7,8,9,10,11,12,13,14], how to generate an array R = [[1,2,3,4], [2,3,4,5], [3,4,5,6], ..., [11,12,13,14]]? (★★★)

    # Author: Stefan van der Walt
    
    Z = np.arange(1,15,dtype=np.uint32)
    R = stride_tricks.as_strided(Z,(11,4),(4,4))
    print(R)
    

    82. Compute a matrix rank (★★★)

    # Author: Stefan van der Walt
    
    Z = np.random.uniform(0,1,(10,10))
    U, S, V = np.linalg.svd(Z) # Singular Value Decomposition
    rank = np.sum(S > 1e-10)
    print(rank)
    

    83. How to find the most frequent value in an array?

    Z = np.random.randint(0,10,50)
    print(np.bincount(Z).argmax())
    

    84. Extract all the contiguous 3x3 blocks from a random 10x10 matrix (★★★)

    # Author: Chris Barker
    
    Z = np.random.randint(0,5,(10,10))
    n = 3
    i = 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)
    

    85. Create a 2D array subclass such that Z[i,j] == Z[j,i] (★★★)

    # Author: Eric O. Lebigot
    # Note: only works for 2d array and value setting using indices
    
    class Symetric(np.ndarray):
        def __setitem__(self, index, value):
            i,j = index
            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] = 42
    print(S)
    

    86. Consider a set of p matrices wich shape (n,n) and a set of p vectors with shape (n,1). How to compute the sum of of the p matrix products at once? (result has shape (n,1)) (★★★)

    # Author: Stefan van der Walt
    
    p, n = 10, 20
    M = 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.
    

    87. Consider a 16x16 array, how to get the block-sum (block size is 4x4)? (★★★)

    # Author: Robert Kern
    
    Z = np.ones((16,16))
    k = 4
    S = 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)
    print(S)
    

    88. How to implement the Game of Life using numpy arrays? (★★★)

    # Author: Nicolas Rougier
    
    def iterate(Z):
        # Count neighbours
        N = (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 rules
        birth = (N==3) & (Z[1:-1,1:-1]==0)
        survive = ((N==2) | (N==3)) & (Z[1:-1,1:-1]==1)
        Z[...] = 0
        Z[1:-1,1:-1][birth | survive] = 1
        return Z
    
    Z = np.random.randint(0,2,(50,50))
    for i in range(100): Z = iterate(Z)
    print(Z)
    

    89. How to get the n largest values of an array (★★★)

    Z = np.arange(10000)
    np.random.shuffle(Z)
    n = 5
    
    # Slow
    print (Z[np.argsort(Z)[-n:]])
    
    # Fast
    print (Z[np.argpartition(-Z,n)[:n]])
    

    90. Given an arbitrary number of vectors, build the cartesian product (every combinations of every item) (★★★)

    # Author: Stefan Van der Walt
    
    def 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).T
    
        for n, arr in enumerate(arrays):
            ix[:, n] = arrays[n][ix[:, n]]
    
        return ix
    
    print (cartesian(([1, 2, 3], [4, 5], [6, 7])))
    

    91. How to create a record array from a regular array? (★★★)

    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')
    print(R)
    

    92. Consider a large vector Z, compute Z to the power of 3 using 3 different methods (★★★)

    # Author: Ryan G.
    
    x = np.random.rand(int(5e7))
    
    %timeit np.power(x,3)
    %timeit x*x*x
    %timeit np.einsum('i,i,i->i',x,x,x)
    

    93. Consider two arrays A and B of shape (8,3) and (2,2). How to find rows of A that contain elements of each row of B regardless of the order of the elements in B? (★★★)

    # Author: Gabe Schwartz
    
    A = np.random.randint(0,5,(8,3))
    B = np.random.randint(0,5,(2,2))
    
    C = (A[..., np.newaxis, np.newaxis] == B)
    rows = np.where(C.any((3,1)).all(1))[0]
    print(rows)
    

    94. Considering a 10x3 matrix, extract rows with unequal values (e.g. [2,2,3]) (★★★)

    # Author: Robert Kern
    
    Z = np.random.randint(0,5,(10,3))
    print(Z)
    # solution for arrays of all dtypes (including string arrays and record arrays)
    E = np.all(Z[:,1:] == Z[:,:-1], axis=1)
    U = Z[~E]
    print(U)
    # soluiton for numerical arrays only, will work for any number of columns in Z
    U = Z[Z.max(axis=1) != Z.min(axis=1),:]
    print(U)
    

    95. Convert a vector of ints into a matrix binary representation (★★★)

    # Author: Warren Weckesser
    
    I = 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. McDonald
    
    I = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128], dtype=np.uint8)
    print(np.unpackbits(I[:, np.newaxis], axis=1))
    

    96. Given a two dimensional array, how to extract unique rows? (★★★)

    # Author: Jaime Fernández del Río
    
    Z = 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)
    
    # Author: Andreas Kouzelis
    # NumPy >= 1.13
    uZ = np.unique(Z, axis=0)
    print(uZ)
    

    97. Considering 2 vectors A & B, write the einsum equivalent of inner, outer, sum, and mul function (★★★)

    # Author: Alex Riley
    # Make sure to read: http://ajcr.net/Basic-guide-to-einsum/
    
    A = np.random.uniform(0,1,10)
    B = np.random.uniform(0,1,10)
    
    np.einsum('i->', A)       # np.sum(A)
    np.einsum('i,i->i', A, B) # A * B
    np.einsum('i,i', A, B)    # np.inner(A, B)
    np.einsum('i,j->ij', A, B)    # np.outer(A, B)
    

    98. Considering a path described by two vectors (X,Y), how to sample it using equidistant samples (★★★)?

    # Author: Bas Swinckels
    
    phi = np.arange(0, 10*np.pi, 0.1)
    a = 1
    x = a*phi*np.cos(phi)
    y = a*phi*np.sin(phi)
    
    dr = (np.diff(x)**2 + np.diff(y)**2)**.5 # segment lengths
    r = np.zeros_like(x)
    r[1:] = np.cumsum(dr)                # integrate path
    r_int = np.linspace(0, r.max(), 200) # regular spaced path
    x_int = np.interp(r_int, r, x)       # integrate path
    y_int = np.interp(r_int, r, y)
    

    99. Given an integer n and a 2D array X, select from X the rows which can be interpreted as draws from a multinomial distribution with n degrees, i.e., the rows which only contain integers and which sum to n. (★★★)

    # Author: Evgeni Burovski
    
    X = 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 = 4
    M = np.logical_and.reduce(np.mod(X, 1) == 0, axis=-1)
    M &= (X.sum(axis=-1) == n)
    print(X[M])
    

    100. Compute bootstrapped 95% confidence intervals for the mean of a 1D array X (i.e., resample the elements of an array with replacement N times, compute the mean of each sample, and then compute percentiles over the means). (★★★)

    # Author: Jessica B. Hamrick
    
    X = np.random.randn(100) # random 1D array
    N = 1000 # number of bootstrap samples
    idx = np.random.randint(0, X.size, (N, X.size))
    means = X[idx].mean(axis=1)
    confint = np.percentile(means, [2.5, 97.5])
    print(confint)
    

    https://github.com/rougier/numpy-100

    相关文章

      网友评论

          本文标题:numpy 100道通关题(三)

          本文链接:https://www.haomeiwen.com/subject/iwhzoctx.html