# Suicide mode on
defaults = np.seterr(all="ignore")
Z = np.ones(1) / 0
# Back to sanity
_ = np.seterr(**defaults)
# 另一個等價的方式， 使用上下文管理器（context manager）
with np.errstate(divide='ignore'):
Z = np.ones(1) / 0
# Author: Evgeni Burovski
X = np.arange(8)
Y = X + 0.5
C = 1.0 / np.subtract.outer(X, Y)
print (C)
print(np.linalg.det(C)) # 計算行列式
48. 列印每個numpy 類型的最小和最大可表示值 (★★☆)
(提示: np.iinfo, np.finfo, eps)
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)
49. 如何列印數組中所有的值？**(★★☆)
(提示: np.set_printoptions)
np.set_printoptions(threshold=np.nan)
Z = np.zeros((16,16))
print(Z)
50. 如何在數組中找到與給定标量接近的值? (★★☆)
(提示: argmin)
Z = np.arange(100)
v = np.random.uniform(0, 100)
index = (np.abs(Z-v)).argmin()
print(Z[index])
A = np.arange(3).reshape(3, 1)
B = np.arange(3).reshape(1, 3)
it = np.nditer([A, B, None])
for x, y, z in it:
z[...] = x + y
print (it.operands[2])
63. 建立一個具有name屬性的數組類 (★★☆)
(提示: class method)
class NameArray(np.ndarray):
def __new__(cls, array, name="no name"):
obj = np.asarray(array).view(cls)
obj.name = name
return obj
def __array_finalize__(self, obj):
if obj is None: return
self.info = getattr(obj, 'name', "no name")
Z = NamedArray(np.arange(10), "range_10")
print (Z.name)
64. 給定一個向量，如何讓在第二個向量索引的每個元素加1(注意重複索引)? (★★★)
(提示: np.bincount | np.add.at)
# 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. 如何根據索引清單
I
将向量
X
的元素累加到數組
F
? (★★★)
(提示: np.bincount)
# 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. 思考(dtype = ubyte)的(w, h, 3)圖像，計算唯一顔色的值**(★★★)
(提示: np.unique)
# 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. 思考如何求一個四維數組最後兩個軸的資料和**(★★★)
(提示: sum(axis=(-2,-1)))
A = np.random.randint(0,10,(3,4,3,4))
# 傳遞一個元組(numpy 1.7.0)
sum = A.sum(axis=(-2,-1))
print(sum)
# 将最後兩個次元壓縮為一個
# (适用于不接受軸元組參數的函數)
sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)
print(sum)
68. 考慮一維向量D，如何使用相同大小的向量S來計算D的子集的均值，其描述子集索引？ (★★★)
(提示: np.bincount)
# 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. **如何獲得點積的對角線？ (★★★)
(提示: np.diag)
# 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)
# 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. 給定一個二進制的數組
C
，如何生成一個數組
A
滿足
np.bincount(A)==C
? (★★★)
(提示: np.repeat)
# 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. 如何通過滑動視窗計算一個數組的平均數? (★★★)
(提示: np.cumsum)
# 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))
# 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. 如何對布爾值取反，或改變浮點數的符号(
sign
)? (★★★)
(提示: np.logical_not, np.negative)
# 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. 思考兩組點集
P0
和
P1
去描述一組線(二維)和一個點
p
,如何計算點
p
到每一條線 i
(P0[i],P1[i])
的距離？ (★★★)
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. 考慮兩組點集
P0
和
P1
去描述一組線(二維)和一組點集
P
，如何計算每一個點
j(P[j])
到每一條線 i
(P0[i],P1[i])
的距離? (★★★)
# 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]))
# 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. 考慮一個數組
Z = [1,2,3,4,5,6,7,8,9,10,11,12,13,14]
,如何生成一個數組
R = [[1,2,3,4], [2,3,4,5], [3,4,5,6], ...,[11,12,13,14]]
? (★★★)
(提示: stride_tricks.as_strided)
# 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. 計算矩陣的秩 (★★★)
(提示: np.linalg.svd)
# 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. 如何找出數組中出現頻率最高的值?**(★★★)
(提示: np.bincount, argmax)
Z = np.random.randint(0,10,50)
print(np.bincount(Z).argmax())
84. 從一個
10x10
的矩陣中提取出連續的
3x3
區塊**(★★★)
(提示: stride_tricks.as_strided)
# 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.建立一個滿足
Z[i,j] == Z[j,i]
的二維數組子類 (★★★)
(提示: class method)
# 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. 考慮p個
nxn
矩陣和一組形狀為
(n,1)
的向量，如何直接計算p個矩陣的乘積
(n,1)
? (★★★)
(提示: np.tensordot)
# 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. 對于一個
16x16
的數組，如何得到一個區域的和(區域大小為
4x4
)? (★★★)
(提示: np.add.reduceat)
# 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. 如何利用
numpy
數組實作Game of Life? (★★★)
(提示: Game of Life , Game of Life有哪些圖形？)
# 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. 如何找到一個數組的第n個最大值?** (★★★)
(提示: np.argsort | np.argpartition)
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. 給定任意個數向量，建立笛卡爾積(每一個元素的每一種組合) (★★★)
(提示: np.indices)
# 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. 如何從一個正常數組中建立記錄數組(
record array
)? (★★★)
(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. 思考一個大向量
Z
, 用三種不同的方法計算它的立方 (★★★)
(提示: np.power, *, np.einsum)
# Author: Ryan G.
x = np.random.rand(5e7)
%timeit np.power(x,3)
%timeit x*x*x
%timeit np.einsum('i,i,i->i',x,x,x)
93. 考慮兩個形狀分别為
(8,3)
和
(2,2)
的數組
A
和
B
. 如何在數組
A
中找到滿足包含
B
中元素的行？(不考慮
B
中每行元素順序)？ (★★★)
(提示: np.where)
# 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. 思考一個
10x3
的矩陣，如何分解出有不全相同值的行 (如
[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. 将一個整數向量轉換為二進制矩陣 (★★★)
(提示: np.unpackbits)
# 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. 給定一個二維數組，如何提取出唯一的行?**(★★★)
(提示: np.ascontiguousarray)
# 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)
97. 考慮兩個向量
A
和
B
，寫出用
einsum
等式對應的
inner, outer, sum, mul
函數 (★★★)
(提示: np.einsum)
# 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. 考慮一個由兩個向量描述的路徑
(X,Y)
，如何用等距樣例(
equidistant samples
)對其進行采樣(
sample
)**(★★★)?
(提示: np.cumsum, np.interp)
# 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)
# 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. 對于一個一維數組
X
，計算它boostrapped之後的95%置信區間的平均值. (★★★)
(提示: np.percentile)
# 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)
