100个Numpy练习【4】

2018-03-01 11:05:58来源:网络收集作者:jzm1995人点击

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接上文: 100个Numpy练习【1】


接上文: 100个Numpy练习【2】


接上文: 100个Numpy练习【3】


Numpy是Python做数据分析必须掌握的基础库之一,非常适合刚学习完Numpy基础的同学,完成以下习题可以帮助你更好的掌握这个基础库。


Python版本:Python 3.6.2


Numpy版本:Numpy 1.13.1





61. 从数组中找出与给定值最接近的值(★★☆)

( 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)
62. 思考形状为(1, 3)和(3, 1)的两个数组形状,如何使用迭代器计算它们的和?(★★☆)

( 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])
63. 创建一个具有name属性的数组类(★★☆)

( 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. 给定一个向量,如何让在第二个向量索引的每个元素加1(注意重复索引)?(★★★)

( 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. 如何根据索引列表 I 将向量 X 的元素累加到数组 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)图像,计算唯一颜色的值(★★★)

( 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. 思考如何求一个四维数组最后两个轴的数据和(★★★)

( 提示 : sum(axis=(-2,-1)))


A = np.random.randint(0,10,(3,4,3,4))
# 传递一个元组(numpy 1.7.0)
sum = A.sum(axis=(-2,-1))
print(sum)
# 将最后两个维度压缩为一个
# (适用于不接受轴元组参数的函数)
sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)
print(sum)
68.考虑一维向量D,如何使用相同大小的向量S来计算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.如何获得点积的对角线?(★★★)

( 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)
70.考虑向量[1,2,3,4,5],如何建立一个新的向量,在每个值之间交错有3个连续的零?(★★★)

( 提示 : array[::4])


# 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. 考虑一个维度(5,5,3)的数组,如何将其与一个(5,5)的数组相乘?(★★★)

( 提示 : array[:, :, None])


A = np.ones((5,5,3))
B = 2*np.ones((5,5))
print(A * B[:,:,None])
72. 如何对一个数组中任意两行做交换?(★★★)

( 提示 : array[[]] = array[[]])


# Author: Eelco Hoogendoorn
A = np.arange(25).reshape(5,5)
A[[0,1]] = A[[1,0]]
print(A)
73. 思考描述10个三角形(共享顶点)的一组10个三元组,找到组成所有三角形的唯一线段集(★★★)

( 提示 : repeat, np.roll, np.sort, view, np.unique)


# 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. 给定一个二进制的数组 C ,如何生成一个数组 A 满足 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. 如何通过滑动窗口计算一个数组的平均数?(★★★)

( 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))
76. 思考以为数组Z,构建一个二维数组,其第一行是(Z[0],Z[1],Z[2]), 然后每一行移动一位,最后一行为 (Z[-3],Z[-2],Z[-1])(★★★)

( 提示 : from numpy.lib import stride_tricks)


# 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. 如何对布尔值取反,或改变浮点数的符号( 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? (★★★)
# 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. 思考一个任意的数组,编写一个函数,该函数提取一个具有固定形状的子部分,并以一个给定的元素为中心(在该部分填充值)(★★★)

( 提示 : minimum, maximum)


# 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)

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