Numpy练习_用python做8×8的棋盘01程序

100道NumPy Exercise题

https://github.com/rougier/numpy-100/blob/master/100_Numpy_exercises.ipynb

• 1.导包查看numpy

import numpy as np
print(np.__version__)
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• 2.创建全0的数组，并查看数组占用内存大小

a=np.zeros(10)
print('%d bytes'%(a.size*a.itemsize))
Z = np.zeros((10,10))
print("%d bytes" % (Z.size * Z.itemsize))
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• 3.创建一个0~49的数组

Z = np.arange(50)
# 修改数组中的值
Z[3:5]=99
print(Z)
# 数组首尾倒序
Z = Z[::-1]
print(Z)
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• 4.数组改变形状reshape

Z = np.arange(9).reshape(3,3)
print(Z)
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• 5.生成10*10的数组，并输出最小和最大值

Z = np.random.random((10,10))
Zmin, Zmax = Z.min(), Z.max()
print(Zmin, Zmax)
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• 6.创建10*10的二维全1数组,让除边界以外的全为0

Z = np.ones((10,10))
# 让除边界以外的全为0
Z[1:-1,1:-1] = 0
print(Z)
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• 7.创建5*5的全1矩阵,设置矩阵的边界为0

Z = np.ones((5,5))
#设置矩阵的边界为0
Z = np.pad(Z, pad_width=1, mode='constant', constant_values=0)
print(Z)
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• 8.创建对角矩阵 对角线上取值1,2,3,4

Z=np.diag((1,2,3,4))
print(Z)
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• 9.设置5*5矩阵对角线以下为1,2,3,4

Z = np.diag(1+np.arange(4),k=-1)
print(Z)
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• 10.创建一个8*8的棋盘状矩阵

Z = np.zeros((8,8),dtype=int)
Z[1::2,::2] = 1 #行数1开始~最后（每次间隔2），列数0开始~最后列（间隔2）
Z[::2,1::2] = 1
print(Z)
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• 11.创建一个8*8的棋盘状矩阵–使用tile对已有的数组进行延展

Z = np.tile( np.array([[0,1],[1,0]]), (4,4))
print(Z)
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• 12.对一个5*5的矩阵进行标准化

Z = np.random.random((5,5))
Z = (Z - np.mean (Z)) / (np.std (Z))
print(Z)
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• 13.矩阵的乘积

Z = np.dot(np.arange(15).reshape(5,3), np.ones((3,4)))
print(Z)

# Alternative solution, in Python 3.5 and above
Z = np.arange(15).reshape(5,3) @ np.ones((3,4))
print(Z)
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• 14.返回一个flatten矩阵的元素的index的横竖索引(x,y)

#numpy.unravel_index(indices, dims, order=‘C’)
#Converts a flat index or array of flat indices into a tuple of coordinate arrays.
# 返回一个flatten矩阵的元素的index的横竖索引(x,y)
print(np.unravel_index(16,(4,5)))# 表示如果把一个4*5的矩阵变成一个扁平的一维数组，索引16在原矩阵中的行坐标和列坐标为3,1
#16=3*5+1
# 返回flat array中索引为100元素在维度（6,7,8）的矩阵中所在的下标
print(np.unravel_index(100,(6,7,8)))
loc_indexs=np.unravel_index([22, 41, 37], (7,6))
print(loc_indexs)
# 22=3*6+4 横竖坐标（3,4）
# 41=6*6+5
# 37=6*6+1
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• 15.Create a custom dtype that describes a color as four unsigned bytes (RGBA)

color = np.dtype([("r", np.ubyte, 1),
                  ("g", np.ubyte, 1),
                  ("b", np.ubyte, 1),
                  ("a", np.ubyte, 1)])
print(color)
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• 16.下面代码结果？

print(sum(range(5),-1))
print(sum(range(5),-3))
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• 17.How to round away from zero a float array ?

# numpy.copysign(x1, x2, /, out=None, *, where=True, casting='same_kind',
#             order='K', dtype=None, subok=True[, signature, extobj])
# Change the sign of x1 to that of x2, element-wise.
Z = np.random.uniform(-10,10,10)
print(Z)
print(np.copysign(np.ceil(np.abs(Z)), Z))
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• 18.数组取交集

Z1 = np.random.randint(0,10,10)
Z2 = np.random.randint(0,10,10)
print(np.intersect1d(Z1,Z2))
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• 19.输出昨天、今天、明天的日期

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:',yesterday,'today:',today,'tomorrow:',tomorrow)
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• 20.判断今天是不是工作日

print(np.is_busday(np.datetime64('today', 'D')))#True
# numpy.busday_count(begindates, enddates, weekmask='1111100', holidays=[], busdaycal=None, out=None)
# 计算两个日期之间的工作日天数
print(np.busday_count('2018-01-01','2018-12-31'))# 260 days
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• 21.获取2019年2月所有的天

Z = np.arange('2019-02', '2019-03', dtype='datetime64[D]')
print(Z)
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• 22.计算((A+B)*(-A/2)) in place (without copy)

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)
print(A)
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• 23.用5种不同的方法提取随机数组中的整数

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))
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• 24.创建一个5x5 矩阵 with row values ranging from 0 to 4

Z = np.zeros((5,5))
Z += np.arange(5)# Z矩阵的每一行都加上[0 1 2 3 4 ]
print(Z)
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• 25.利用生成器产生10个整数，组成数组

def generate():
    for x in range(10):
        yield x
Z = np.fromiter(generate(),dtype=float,count=-1)
print(Z)
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• 26.创建一个从0~1之间的10个数，不包括两端端点

Z = np.linspace(0,1,11,endpoint=False)[1:]
print(Z)
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• 27.样对一个小的数组进行累加求和，要求比np.sum更快

Z = np.arange(10)
np.add.reduce(Z)
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• 28.比较两个矩阵的相似性

A = np.random.randint(0,2,5)
B = np.random.randint(0,2,5)

# 假定两个矩阵相同形状，可以容忍相似度存在一定的差异
# numpy.allclose(a, b, rtol=1e-05, atol=1e-08, equal_nan=False)[source]
# Returns True if two arrays are element-wise equal within a tolerance.
equal = np.allclose(A,B)
print(equal)

# 确定的比较两个矩阵的形状和对值进行逐一比较，零容忍：要么完全相同，要么不同
equal = np.array_equal(A,B)
print(equal)
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• 29.创建一个不可修改的数组，只读

Z = np.zeros(10)
Z.flags.writeable = False#设置不可修改
Z[0] = 1#对数组进行修改，会报错
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• 30.创建一个10x2的矩阵（默认直角坐标系-笛卡尔坐标系），转换成极坐标系–牛顿创立

Z = np.random.random((10,2))
X,Y = Z[:,0], Z[:,1]
R = np.sqrt(X**2+Y**2)# 极坐标系半径
T = np.arctan2(Y,X)#旋转角度
print(R)
print(T)
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• 31.创建一个10个数的数组，将最大值替换成0

Z = np.random.random(10)
print(Z)
Z[Z.argmax()] = 0#argmax:返回最大值所在下标index，argmin返回最小值的index
print(Z)
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• 32.创建（x,y）坐标，覆盖[0,1]x[0,1]的正方形区域

Z = np.zeros((5,5), [('x',float),('y',float)])#自定义dtype
print(Z)
print(Z['x'])
Z['x'], Z['y'] = np.meshgrid(np.linspace(0,1,5),
                             np.linspace(0,1,5))
print(Z['y'])#y横向排列，每一行中数值相同，x纵向排列，每一列中数值相同
print(Z)
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• 32.构建柯西矩阵Cauchy matrix C (Cij =1/(xi - yj))

X = np.arange(8)
Y = X + 0.5
C = 1.0 / np.subtract.outer(X, Y)
print(np.linalg.det(C))#计算矩阵行列式
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• 33.打印numpy中标量类型所能表示的最大值和最小值

for dtype in [np.int8, np.int32, np.int64]:
   print(str(dtype)+'min:',np.iinfo(dtype).min)
   print(str(dtype)+'max:',np.iinfo(dtype).max)
# int8[-128,127]
# int32[-2147483648,2147483647]
# int32[-9223372036854775808,9223372036854775807]  

for dtype in [np.float32, np.float64]:
   print(str(dtype)+'min:',np.finfo(dtype).min)
   print(str(dtype)+'max:',np.finfo(dtype).max)
   print(np.finfo(dtype).eps)#float所能表示的精度
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• 34.打印数组中的所有的值

np.set_printoptions(threshold=np.nan)
Z = np.zeros((16,16))
print(Z)
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• 35.从一个数组中找出与给定标量值最相近的值

Z = np.arange(100)
v = np.random.uniform(0,100)
print('v:',v)
index = (np.abs(Z-v)).argmin()
print(Z[index])
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• 36.Create a structured array representing a position (x,y) and a color (r,g,b)

Z = np.zeros(10, [ ('position', [('x', float, 1),('y', float, 1)]),
                   ('color',    [('r', float, 1),('g', float, 1),('b', float, 1)])
                 ]       
            )
print(Z)
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• 37.100x2维的矩阵，计算点与点之间的距离,类似于求两两之间的相似度

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)
print('-'*50)
# Much faster with scipy
import scipy
import scipy.spatial
Z = np.random.random((10,2))
D = scipy.spatial.distance.cdist(Z,Z)
print(D)
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• 38.把float32的数组转成int32的数组，without copy

Z = np.arange(10, dtype=np.float32)
Z = Z.astype(np.int32, copy=False)
print(Z)
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• 39.np读取文件数据

from io import StringIO

# Fake file 
s = StringIO("""1, 2, 3, 4, 5\n
                6,  ,  , 7, 8\n
                 ,  , 9,10,11\n""")
Z = np.genfromtxt(s, delimiter=",", dtype=np.int)
print(Z
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• 40.np.ndenumerate与Python中的enumerate相类似

Z = np.arange(9).reshape(3,3)
for index, value in np.ndenumerate(Z):
    print(index, value)
print('-'*50)

for index in np.ndindex(Z.shape):
    print(index, Z[index])
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• 41.生成2D的高斯分布数组

X, Y = np.meshgrid(np.linspace(-1,1,5), np.linspace(-1,1,5))
print(X)
print(Y)
D = np.sqrt(X*X+Y*Y)
sigma, mu = 1.0, 0.0
G = np.exp(-( (D-mu)**2 / ( 2.0 * sigma**2 ) ) )#正态分布概率密度函数
print(G)
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• 42.np.put随机把p个元素放置在2D array中

n = 10
p = 3
Z = np.zeros((n,n))
# numpy.put(a, ind, v, mode='raise')
# Replaces specified elements of an array with given values.
np.put(Z, np.random.choice(range(n*n), p, replace=False),2)
print(Z)
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• 43.矩阵的每一行减去该行的均值

X = np.random.rand(3, 5)#3x5的矩阵

# Recent versions of numpy
Y = X - X.mean(axis=1, keepdims=True)
# 或者
# Older versions of numpy
Y = X - X.mean(axis=1).reshape(-1, 1)

print(Y)
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• 44.对数组按照指定列进行排序

Z = np.random.randint(0,10,(3,3))
print(Z)
print(Z[Z[:,1].argsort()])#对数组按照第2列进行排序
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• 45.判断矩阵是否存在空列

Z = np.random.randint(0,3,(3,10))
print(Z)
# np.array.any()是或操作，将np.array中所有元素进行或操作，然后返回True或False
# np.array.all()是与操作，将np.array中所有元素进行与操作，然后返回True或False
print((~Z.any(axis=0)).any())
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• 46.找出数组中与给定值最相近的数

Z = np.random.uniform(0,1,10)
print(Z)
z = 0.5
# ndarray.flat 将数组转换为1-D的迭代器 / 
# flat返回的是一个迭代器，可以用for访问数组每一个元素

m = Z.flat[np.abs(Z - z).argmin()]
print(m)

# ndarray.flatten(order=’C’)
# Return a copy of the array collapsed into one dimension. 
# 将数组的副本转换为一个维度，并返回
# order：{‘C’,‘F’,‘A’,‘K’}
# 可选参数，order：{‘C’,‘F’,‘A’,‘K’}
# ‘C’：C-style，行序优先
# ‘F’：Fortran-style，列序优先
# ‘A’：if a is Fortran contiguous in memory ,flatten in column_major order
# ‘K’：按照元素在内存出现的顺序进行排序 
# 默认为’C’
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• 46.运用迭代器计算13数组和31数组的和

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])
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• 47.np.nditer: numpy array自带的迭代器

# it = np.nditer(x, flags=['multi_index'], op_flags=['readwrite'])
# numpy array自带的迭代器，multi_index：多重索引[输出：就是数组索引号（0,0）（0,1）等]; 
# readwrite：对x有可读可写的权限;iternext()：控制下一次迭代，如果没有这个控制下一次，就一直在一个索引值上操作;挺好用。
a=np.arange(0,60,5).reshape(3,4)
it=np.nditer(a,flags=['multi_index'],op_flags=['readwrite'])
while not it.finished:
    print(it.multi_index)#输出迭代器索引
    it.iternext()
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• 48.Create an array class that has a name attribute (★★☆)

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

Z = np.ones(10)
# numpy.random.randint
# low、high、size三个参数。默认high是None,如果只有low，那范围就是[0,low)。如果有high，范围就是[low,high)。
I = np.random.randint(0,len(Z),20)
Z += np.bincount(I, minlength=len(Z))
print(Z)
# 关于numpy.bincount详解：https://blog.csdn.net/xlinsist/article/details/51346523
# bin的数量比x中的最大值大1，每个bin给出了它的索引值在x中出现的次数

# Another solution
np.add.at(Z, I, 1)
print(Z)
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• 50.How to accumulate elements of a vector (X) to an array (F) based on an index list (I)? (★★★)

X = [1,2,3,4,5,6]
I = [1,3,9,3,4,1]
F = np.bincount(I,X)#X为权重weight数组，out[n] += weight[i]
print(F)
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• 51.Considering a (w,h,3) image of (dtype=ubyte), compute the number of unique colors (★★★)

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)
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• 52.四维矩阵，通过最后两个轴快速求和

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

D = np.random.uniform(0,1,100)# 均匀分布[0,1),左闭右开，size=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())
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• 54.获取矩阵点乘后的对角线元素值

A = np.random.uniform(0,1,(5,5))
B = np.random.uniform(0,1,(5,5))

# np.diag(array)
# array是一个1维数组时，结果形成一个以一维数组为对角线元素的矩阵
# array是一个二维矩阵时，结果输出矩阵的对角线元素

# 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)
# numpy.einsum(subscripts, *operands, out=None, dtype=None, order='K', casting='safe', optimize=False)
# subscripts用于指定计算模式,operands用于指定操作数
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• 55.在向量[1, 2, 3, 4, 5]的每个值之间放3个0

Z = np.array([1,2,3,4,5])
nz = 3
Z0 = np.zeros(len(Z) + (len(Z)-1)*(nz))
Z0[::nz+1] = Z#把Z中原来的值放入Z0中，每间隔4
print(Z0)
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• 56.一个553的数组如何与5*5的数组相乘

A = np.ones((5,5,3))
B = 2*np.ones((5,5))
print(A * B[:,:,None])#由于维度不够，在第三维上使用None填充
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• 57.交换数组的两行

A = np.arange(25).reshape(5,5)
A[[0,1]] = A[[1,0]]
print(A)
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• 58.交换数组的两列

A = np.arange(25).reshape(5,5)
A[:,[0,1]]=A[:,[1,0]]
print(A)
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• 59.Consider a set of 10 triplets describing 10 triangles (with shared vertices), find the set of unique line segments composing all the triangles (★★★)

faces = np.random.randint(0,100,(10,3))
#  np.roll(a, shift, axis=None)将a，沿着axis的方向，滚动shift长度
# numpy.repeat(a, repeats, axis=None)[source]
# Repeat elements of an array.
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)
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• 60.Given an array C that is a bincount, how to produce an array A such that np.bincount(A) == C? (★★★)

C = np.bincount([1,1,2,3,4,4,6])
print(np.arange(len(C)))
print(C)
A = np.repeat(np.arange(len(C)), C)
print(A)
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• 61.在数组中利用滑动窗口计算移动平均值 (★★★)

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))
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• 62.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]) (★★★)

from numpy.lib import stride_tricks
# as_strided creates a view into the array given the exact strides and shape. 
# This means it manipulates the internal data structure of ndarray and, 
# if done incorrectly, the array elements can point to invalid memory and can corrupt results or crash your program. 
# itemsize输出array元素的字节数：int：4,float：8
def rolling(a, window):
    shape = (a.size - window + 1, window)
    strides = (a.itemsize, a.itemsize)#（4,4）
    return stride_tricks.as_strided(a, shape=shape, strides=strides)
Z = rolling(np.arange(10), 3)
print(Z)
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• 63.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]]? (★★★)

from numpy.lib import stride_tricks
Z = np.arange(1,15,dtype=np.uint32)
R = stride_tricks.as_strided(Z,(11,4),(4,4))
print(R)
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• 64.Extract all the contiguous 3x3 blocks from a random 10x10 matrix (★★★)

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)
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• 64.boolean型数组取反，或者数组对元素取相反数

import numpy as np
Z = np.random.randint(0,2,100)
np.logical_not(Z, out=Z)

Z = np.random.uniform(-1.0,1.0,20)
np.negative(Z, out=Z)#对数组元素取反
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• 65.计算点p到各条直线（p0,p1）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))
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• 66.计算多个点到多条直线的距离

# 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, (5,2))#5个点
print(np.array([distance(P0,P1,p_i) for p_i in p]))
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• 67.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) (★★★)

Z = np.random.randint(0,10,(10,10))
shape = (5,5)
fill  = 0
position = (1,1)#以index（1,1）为中心，从Z矩阵中抽取一个5*5的矩阵，不足部分填充0

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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• 68.矩阵分解、计算矩阵的秩

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)
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• 69.在数组中找出出现次数最多的值

Z = np.random.randint(0,10,50)
print(np.bincount(Z).argmax())
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• 70.创建一个对称的矩阵

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)
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• 71.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)) (★★★)

p, n = 10, 20
M = np.ones((p,n,n))
V = np.ones((p,n,1))
# 张量乘积tensordot()将两个多维数组a和b指定轴上的对应元素相乘并求和，它是最一般化的乘积运算函数。
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.
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• 72.Consider a 16x16 array, how to get the block-sum (block size is 4x4)? (★★★)

Z = np.ones((16,16))
k = 4
# 先横着把行的四个相加，在把竖着列相加
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)
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• 73.How to implement the Game of Life using numpy arrays? (★★★)

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)
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• 74.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]])
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• 75.给定多个数组，构建一个笛卡尔积矩阵（元素之间两两组合） (★★★)

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])))
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• 76.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)
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[(b’Hello’, 2.5, 3) (b’World’, 3.6, 2)]

• 76.计算一个超大向量的三次方

# np.random.rand返回一个或一组服从“0~1”均匀分布的随机样本值。随机样本取值范围是[0,1)，不包括1。
x = np.random.rand(5000)#生成5*10^7个[0,1）之间的数

# timeit准确测量小段代码的执行时间
# ipython中以%开头的叫做line magic, 这种类型的指令只能作用于一行代码，默认是可以不带百分号使用的。
# 以%%开头的叫做cell magic, 这种类型的指令只能作用于代码块。
%timeit np.power(x,3)
%timeit x*x*x
%timeit np.einsum('i,i,i->i',x,x,x)
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748 µs ± 53.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
22 µs ± 3.77 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
30.2 µs ± 2.17 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

• 77.A矩阵83，B矩阵22，找出A矩阵中包含所有B矩阵值的行，不考虑顺序 (★★★)

A = np.random.randint(0,5,(8,3))
B = np.random.randint(0,5,(2,2))
print(A)
print(B)

C = (A[..., np.newaxis, np.newaxis] == B)
rows = np.where(C.any((3,1)).all(1))[0]
print('rows:',rows)
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• 78.Considering a 10x3 matrix, extract rows with unequal values (e.g. [2,2,3]) (★★★)

Z = np.random.randint(0,4,(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)
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• 79.Convert a vector of ints into a matrix binary representation (★★★)

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])

# another 方法

I = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128], dtype=np.uint8)
print(np.unpackbits(I[:, np.newaxis], axis=1))
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• 80.Given a two dimensional array, how to extract unique rows? (★★★)

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

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

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)
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• 83.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. (★★★)

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])
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[[2. 0. 1. 1.]]

• 84.# 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). (★★★)

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)
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[-0.20332504 0.15369648]