后面的题目基本都写不出来，平时遇到的也不多，为了文章的完整性，将答案贴出。
- 考虑一维数组Z，构建一个二维数组， 其第一行是（Z [0]，Z [1]，Z [2]），每个后续行移1（最后一行应该是（ Z [-3]，Z [-2]，Z [-1]）(提示: from numpy.lib import stride_tricks)
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)
- 如何对布尔值取反，或者原位(in-place)改变浮点数的符号(sign)？(提示: np.logical_not, np.negative)
# 方法一
Z = np.random.randint(0, 2, 100)
np.logical_not(Z, out=Z)
# 方法二
Z = np.random.uniform(-1.0, 1.0, 100)
np.negative(Z, out=Z)
- 考虑两组点集P0和P1去描述一组线(二维)和一个点p,如何计算点p到每一条线 i (P0[i],P1[i])的距离？
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.考虑两组点集P0和P1去描述一组线(二维)和一组点集P，如何计算每一个点j(P[j])到每一条线i(P0[i], P1[i])的距离？
# 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.考虑一个任意数组，写一个函数，提取一个固定形状的子部分，并以给定元素为中心（fill必要时填充一个值）(提示: minimum, maximum)
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.考虑一个数组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]]?(提示: stride_tricks.as_strided)
Z = np.arange(1, 15, dtype=np.uint32)
R = stride_tricks.as_strided(Z, (11, 4), (4, 4))
print(R)
82.计算一个矩阵的秩(提示: np.linalg.svd)
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.如何找到一个数组中出现频率最高的值？(提示: np.bincount, argmax)
Z = np.random.randint(0, 10, 50)
print(np.bincount(Z).argmax())
84.从一个10x10的矩阵中提取出连续的3x3区块(提示: stride_tricks.as_strided)
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.创建一个满足Z[i, j] == Z[j, i]的子类(提示:class方法)
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.考虑p个nxn矩阵和一组形状为(n, 1)的向量，如何直接计算p个矩阵的乘积(n, 1)？(提示: np.tensordot)
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.对于一个16x16的数组，如何得到一个区域(block - sum)的和(区域大小为4x4)?(提示: np.add.reduceat)
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.如何利用numpy数组实现GameofLife?(提示: Game of Life)
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.如何找到一个数组的第n个最大值?(提示: np.argsort | np.argpartition)
Z = np.arange(10000)
np.random.shuffle(Z)
n = 5
# 方法1
# Slow
print(Z[np.argsort(Z)[-n:]])
# 方法2
# Fast
print(Z[np.argpartition(-Z, n)[:n]])
90.给定任意个数向量，创建笛卡尔积(每一个元素的每一种组合)(提示: np.indices)
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.如何从一个正常数组创建记录数组(recordarray)?(提示: np.core.records.fromarrays)
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.考虑一个大向量Z, 用三种不同的方法计算它的立方(提示: np.power, \ *, np.einsum)
# 方法1
x = np.random.rand()
np.power(x, 3)
# 方法2
x * x * x
# 方法3
np.einsum('i,i,i->i', x, x, x)
93.考虑两个形状分别为(8, 3)和(2, 2)的数组A和B.如何在数组A中找到满足包含B中元素的行？(不考虑B中每行元素顺序)？(提示: np.where)
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.考虑一个10x3的矩阵，分解出有不全相同值的行(如[2, 2, 3])
Z = np.random.randint(0, 5, (10, 3))
print(Z)
# 方法1
# 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)
# 方法2
# 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.将一个整数向量转换为matrixbinary的表现形式(提示: np.unpackbits)
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])
# 方法2
print(np.unpackbits(I[:, np.newaxis], axis=1))
96.给定一个二维数组，如何提取出唯一的(unique)行？(提示: np.ascontiguousarray)
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)
97.考虑两个向量A和B，写出用einsum等式对应的inner, outer, sum, mul函数(提示: np.einsum)
A = np.random.uniform(0, 1, 10)
B = np.random.uniform(0, 1, 10)
print('sum')
print(np.einsum('i->', A)) # np.sum(A)
print('A * B')
print(np.einsum('i,i->i', A, B)) # A * B
print('inner')
print(np.einsum('i,i', A, B)) # np.inner(A, B)
print('outer')
print(np.einsum('i,j->ij', A, B)) # np.outer(A, B)
98.考虑一个由两个向量描述的路径(X, Y)，如何用等距样例(equidistantsamples)对其进行采样(sample)?(提示: np.cumsum, np.interp)
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.给定整数n和2D数组X，从X中选择可以解释为具有n度的多项分布的绘制的行，即，仅包含整数并且总和为n的行。(提示: np.logical_and.reduce, np.mod)
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.计算1D阵列X的平均值的自举95％置信区间（即，对替换N次的阵列的元素进行重新采样，计算每个样本的平均值，然后计算均值上的百分位数）。
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)
网友评论