Datawhale,和鲸社区编辑
Numpy是Python做数据分析所必须要掌握的基础库之一。以下为入门Numpy的100题小练习,原为github上的开源项目,由和鲸社区的小科翻译并整理(保留了部分原文作为参考)。受限于篇幅,小编在这里只提供了部分题目的运行结果。友情提示:代码虽好,自己动手才算学到。
未经授权,禁止转载。
1. 导入numpy库并简写为 np (★☆☆)
(提示: import … as …)
import numpy as np
(提示: np.version, np.show_config)
print(np.__version__)
np.show_config()
(提示: np.zeros)
Z = np.zeros(10)
print(Z)
(提示: size, itemsize)
Z = np.zeros((10,10))
print("%d bytes" % (Z.size * Z.itemsize))
(提示: np.info)
numpy.info(numpy.add)
add(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj])
(提示: array[4])
Z = np.zeros(10)
Z[4] = 1
print(Z)
(提示: np.arange)
Z = np.arange(10,50)
print(Z)
(提示: array[::-1])
Z = np.arange(50)
Z = Z[::-1]
print(Z)
(提示: reshape)
Z = np.arange(9).reshape(3,3)
print(Z)
(提示: np.nonzero)
nz = np.nonzero([1,2,0,0,4,0])
print(nz)
(提示: np.eye)
Z = np.eye(3)
print(Z)
(提示: np.random.random)
Z = np.random.random((3,3,3))
print(Z)
(提示: min, max)
Z = np.random.random((10,10))
Zmin, Zmax = Z.min(), Z.max()
print(Zmin, Zmax)
14. 创建一个长度为30的随机向量并找到它的平均值 (★☆☆)
(提示: mean)
Z = np.random.random(30)
m = Z.mean()
print(m)
(提示: array[1:-1, 1:-1])
Z = np.ones((10,10))
Z[1:-1,1:-1] = 0
print(Z)
(提示: np.pad)
Z = np.ones((5,5))
Z = np.pad(Z, pad_width=1, mode='constant', constant_values=0)
print(Z)
(提示: NaN = not a number, inf = infinity)
0 * np.nan np.nan == np.nan np.inf > np.nan np.nan - np.nan 0.3 == 3 * 0.1
print(0 * np.nan)
print(np.nan == np.nan)
print(np.inf > np.nan)
print(np.nan - np.nan)
print(0.3 == 3 * 0.1)
18. 创建一个 5x5的矩阵,并设置值1,2,3,4落在其对角线下方位置 (★☆☆)
(提示: np.diag)
Z = np.diag(1+np.arange(4),k=-1)
print(Z)
(提示: array[::2])
Z = np.zeros((8,8),dtype=int)
Z[1::2,::2] = 1
Z[::2,1::2] = 1
print(Z)
20. 考虑一个 (6,7,8) 形状的数组,其第100个元素的索引(x,y,z)是什么?
(提示: np.unravel_index)
print(np.unravel_index(100,(6,7,8)))
(提示: np.tile)
Z = np.tile( np.array([[0,1],[1,0]]), (4,4))
print(Z)
(提示: (x - min) / (max - min))
Z = np.random.random((5,5))
Zmax, Zmin = Z.max(), Z.min()
Z = (Z - Zmin)/(Zmax - Zmin)
print(Z)
(提示: np.dtype)
color = np.dtype([("r", np.ubyte, 1),
("g", np.ubyte, 1),
("b", np.ubyte, 1),
("a", np.ubyte, 1)])
color
(提示: np.dot | @)
Z = np.dot(np.ones((5,3)), np.ones((3,2)))
print(Z)
25. 给定一个一维数组,对其在3到8之间的所有元素取反 (★☆☆)
(提示: >, <=)
Z = np.arange(11)
Z[(3 < Z) & (Z <= 8)] *= -1
print(Z)
26. 下面脚本运行后的结果是什么? (★☆☆)
(提示: np.sum)
print(sum(range(5),-1)) from numpy import * print(sum(range(5),-1))
print(sum(range(5),-1))
from numpy import *
print(sum(range(5),-1))
Z**Z 2 << Z >> 2 Z <- Z 1j*Z Z/1/1 ZZ
Z = np.arange(5)
Z ** Z # legal
array([ 1, 1, 4, 27, 256])
Z = np.arange(5)
2 << Z >> 2 # false
array([0, 1, 2, 4, 8])
Z = np.arange(5)
Z <- Z # legal
array([False, False, False, False, False])
Z = np.arange(5)
1j*Z # legal
array([0.+0.j, 0.+1.j, 0.+2.j, 0.+3.j, 0.+4.j])
Z = np.arange(5)
Z/1/1 # legal
array([0., 1., 2., 3., 4.])
Z = np.arange(5)
Z<Z>Z # false
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
np.array(0) / np.array(0) np.array(0) // np.array(0) np.array([np.nan]).astype(int).astype(float)
print(np.array(0) / np.array(0))
print(np.array(0) // np.array(0))
print(np.array([np.nan]).astype(int).astype(float))
(提示: np.uniform, np.copysign, np.ceil, np.abs)
Z = np.random.uniform(-10,+10,10)
print (np.copysign(np.ceil(np.abs(Z)), Z))
(提示: np.intersect1d)
Z1 = np.random.randint(0,10,10)
Z2 = np.random.randint(0,10,10)
print(np.intersect1d(Z1,Z2))
(提示: np.seterr, np.errstate)
# Suicide mode on
defaults = np.seterr(all="ignore")
Z = np.ones(1) / 0
# Back to sanity
_ = np.seterr(**defaults)
An equivalent way, with a context manager:
with np.errstate(divide='ignore'):
Z = np.ones(1) / 0
(提示: imaginary number)
np.sqrt(-1) == np.emath.sqrt(-1)
np.sqrt(-1) == np.emath.sqrt(-1)
False
(提示: np.datetime64, np.timedelta64)
yesterday = np.datetime64('today', 'D') - np.timedelta64(1, 'D')
today = np.datetime64('today', 'D')
tomorrow = np.datetime64('today', 'D') + np.timedelta64(1, 'D')
print ("Yesterday is " + str(yesterday))
print ("Today is " + str(today))
print ("Tomorrow is "+ str(tomorrow))
(提示: np.arange(dtype=datetime64['D']))
Z = np.arange('2016-07', '2016-08', dtype='datetime64[D]')
print(Z)
(提示: np.add(out=), np.negative(out=), np.multiply(out=), np.divide(out=))
A = np.ones(3)*1
B = np.ones(3)*2
C = np.ones(3)*3
np.add(A,B,out=B)
np.divide(A,2,out=A)
np.negative(A,out=A)
np.multiply(A,B,out=A)
array([-1.5, -1.5, -1.5])
(提示: %, np.floor, np.ceil, astype, np.trunc)
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))
(提示: np.arange)
Z = np.zeros((5,5))
Z += np.arange(5)
print (Z)
(提示: np.fromiter)
def generate():
for x in range(10):
yield x
Z = np.fromiter(generate(),dtype=float,count=-1)
print (Z)
[0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]
(提示: np.linspace)
Z = np.linspace(0,1,11,endpoint=False)[1:]
print (Z)
40. 创建一个长度为10的随机向量,并将其排序 (★★☆)
(提示: sort)
Z = np.random.random(10)
Z.sort()
print (Z)
(提示: np.add.reduce)
Z = np.arange(10)
np.add.reduce(Z)
(提示: np.allclose, np.array_equal)
A = np.random.randint(0,2,5)
B = np.random.randint(0,2,5)
# Assuming identical shape of the arrays and a tolerance for the comparison of values
equal = np.allclose(A,B)
print(equal)
False
# 方法2
# Checking both the shape and the element values, no tolerance (values have to be exactly equal)
equal = np.array_equal(A,B)
print(equal)
False
(提示: flags.writeable)
# 使用如下过程实现
Z = np.zeros(10)
Z.flags.writeable = False
Z[0] = 1
44. 将笛卡尔坐标下的一个10x2的矩阵转换为极坐标形式(★★☆)
(hint: np.sqrt, np.arctan2)
Z = np.random.random((10,2))
X,Y = Z[:,0], Z[:,1]
R = np.sqrt(X**2+Y**2)
T = np.arctan2(Y,X)
print (R)
print (T)
(提示: argmax)
Z = np.random.random(10)
Z[Z.argmax()] = 0
print (Z)
(提示: np.meshgrid)
Z = np.zeros((5,5), [('x',float),('y',float)])
Z['x'], Z['y'] = np.meshgrid(np.linspace(0,1,5),
np.linspace(0,1,5))
print(Z)
(提示: np.subtract.outer)
X = np.arange(8)
Y = X + 0.5
C = 1.0 / np.subtract.outer(X, Y)
print(np.linalg.det(C))
(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)
(提示: np.set_printoptions)
np.set_printoptions(threshold=np.nan)
Z = np.zeros((16,16))
print (Z)
(提示: argmin)
Z = np.arange(100)
v = np.random.uniform(0,100)
index = (np.abs(Z-v)).argmin()
print (Z[index])
(提示: dtype)
Z = np.zeros(10, [ ('position', [ ('x', float, 1),
('y', float, 1)]),
('color', [ ('r', float, 1),
('g', float, 1),
('b', float, 1)])])
print (Z)
(提示: np.atleast_2d, T, np.sqrt)
Z = np.random.random((10,2))
X,Y = np.atleast_2d(Z[:,0], Z[:,1])
D = np.sqrt( (X-X.T)**2 + (Y-Y.T)**2)
print (D)
# 方法2
# Much faster with scipy
import scipy
# Thanks Gavin Heverly-Coulson (#issue 1)
import scipy.spatial
D = scipy.spatial.distance.cdist(Z,Z)
print (D)
(提示: astype(copy=False))
Z = np.arange(10, dtype=np.int32)
Z = Z.astype(np.float32, copy=False)
print (Z)
(提示: np.genfromtxt)
1, 2, 3, 4, 5
6, , , 7, 8
, , 9,10,11
参考链接:https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.genfromtxt.html
(提示: np.ndenumerate, np.ndindex)
Z = np.arange(9).reshape(3,3)
for index, value in np.ndenumerate(Z):
print (index, value)
for index in np.ndindex(Z.shape):
print (index, Z[index])
56. 生成一个通用的二维Gaussian-like数组 (★★☆)
(提示: np.meshgrid, np.exp)
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, 0.0
G = np.exp(-( (D-mu)**2 / ( 2.0 * sigma**2 ) ) )
print (G)
(提示: np.put, np.random.choice)
n = 10
p = 3
Z = np.zeros((n,n))
np.put(Z, np.random.choice(range(n*n), p, replace=False),1)
print (Z)
(提示: mean(axis=,keepdims=))
X = np.random.rand(5, 10)
# Recent versions of numpy
Y = X - X.mean(axis=1, keepdims=True)
print(Y)
# 方法2
# Older versions of numpy
Y = X - X.mean(axis=1).reshape(-1, 1)
print (Y)
(提示: argsort)
Z = np.random.randint(0,10,(3,3))
print (Z)
print (Z[Z[:,1].argsort()])
(提示: any, ~)
Z = np.random.randint(0,3,(3,10))
print ((~Z.any(axis=0)).any())
True
(提示: np.abs, argmin, flat)
Z = np.random.uniform(0,1,10)
z = 0.5
m = Z.flat[np.abs(Z - z).argmin()]
print (m)
0.5531249196891759
(提示: np.nditer)
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])
(提示: class方法)
class NamedArray(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)
range_10
(提示: np.bincount | np.add.at)
Z = np.ones(10)
I = np.random.randint(0,len(Z),20)
Z += np.bincount(I, minlength=len(Z))
print(Z)
[3. 1. 5. 4. 3. 4. 2. 1. 4. 3.]
# 方法2
np.add.at(Z, I, 1)
print(Z)
[5. 1. 9. 7. 5. 7. 3. 1. 7. 5.]
(提示: np.bincount)
X = [1,2,3,4,5,6]
I = [1,3,9,3,4,1]
F = np.bincount(I,X)
print (F)
[0. 7. 0. 6. 5. 0. 0. 0. 0. 3.]
(提示: np.unique)
w,h = 16,16
I = np.random.randint(0,2,(h,w,3)).astype(np.ubyte)
#Note that we should compute 256*256 first.
#Otherwise numpy will only promote F.dtype to 'uint16' and overfolw will occur
F = I[...,0]*(256*256) + I[...,1]*256 +I[...,2]
n = len(np.unique(F))
print (n)
8
(提示: sum(axis=(-2,-1)))
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)
# 方法2
sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)
print (sum)
(提示: np.bincount)
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)
# 方法2
import pandas as pd
print(pd.Series(D).groupby(S).mean())
(提示: np.diag)
A = np.random.uniform(0,1,(5,5))
B = np.random.uniform(0,1,(5,5))
# slow version
np.diag(np.dot(A, B))
# 方法2
# Fast version
np.sum(A * B.T, axis=1)
# 方法3
# Faster version
np.einsum("ij,ji->i", A, B)
(提示: array[::4])
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)
[1. 0. 0. 0. 2. 0. 0. 0. 3. 0. 0. 0. 4. 0. 0. 0. 5.]
(提示: array[:, :, None])
A = np.ones((5,5,3))
B = 2*np.ones((5,5))
print (A * B[:,:,None])
(提示: array[[]] = array[[]])
A = np.arange(25).reshape(5,5)
A[[0,1]] = A[[1,0]]
print (A)
73. 考虑一个可以描述10个三角形的triplets,找到可以分割全部三角形的line segment
Consider a set of 10 triplets describing 10 triangles (with shared vertices), find the set of unique line segments composing all the triangles (★★★) (提示: repeat, np.roll, np.sort, view, np.unique)
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. 给定一个二进制的数组C,如何产生一个数组A满足np.bincount(A)==C(★★★)
(提示: np.repeat)
C = np.bincount([1,1,2,3,4,4,6])
A = np.repeat(np.arange(len(C)), C)
print (A)
[1 1 2 3 4 4 6]
(提示: np.cumsum)
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))
[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.]
(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)
(提示: np.logical_not, np.negative)
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)
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))
# 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]))
(hint: 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)
(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)
(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)
(提示: np.bincount, argmax)
Z = np.random.randint(0,10,50)
print (np.bincount(Z).argmax())
1
(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)
(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)
(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)
(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)
(提示: np.argsort | np.argpartition)
Z = np.arange(10000)
np.random.shuffle(Z)
n = 5
# Slow
print (Z[np.argsort(Z)[-n:]])
[9995 9996 9997 9998 9999]
# 方法2
# Fast
print (Z[np.argpartition(-Z,n)[:n]])
[9999 9997 9998 9996 9995]
(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])))
(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)
[(b'Hello', 2.5, 3) (b'World', 3.6, 2)]
(提示: np.power, *, np.einsum)
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)
(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)
[0 1 4 5 6 7]
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)
# 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)
(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))
(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)
Considering a path described by two vectors (X,Y), how to sample it using equidistant samples (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)
(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])
[[2. 0. 1. 1.]]
(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). (★★★) (提示: np.percentile)
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)
原文地址:https://www.kesci.com/home/project/59f29f67c5f3f5119527a2cc