Numpy基础练习100题

项目来源：https://github.com/rougier/numpy-100

1.导入numpy库并取名字为np

import numpy as np

2.打印输出numpy的版本和配置信息

np.__version__

'1.18.5'

np.show_config()

blas_mkl_info:NOT AVAILABLE
blis_info:NOT AVAILABLE
openblas_info:library_dirs = ['C:\\projects\\numpy-wheels\\numpy\\build\\openblas_info']libraries = ['openblas_info']language = f77define_macros = [('HAVE_CBLAS', None)]
blas_opt_info:library_dirs = ['C:\\projects\\numpy-wheels\\numpy\\build\\openblas_info']libraries = ['openblas_info']language = f77define_macros = [('HAVE_CBLAS', None)]
lapack_mkl_info:NOT AVAILABLE
openblas_lapack_info:library_dirs = ['C:\\projects\\numpy-wheels\\numpy\\build\\openblas_lapack_info']libraries = ['openblas_lapack_info']language = f77define_macros = [('HAVE_CBLAS', None)]
lapack_opt_info:library_dirs = ['C:\\projects\\numpy-wheels\\numpy\\build\\openblas_lapack_info']libraries = ['openblas_lapack_info']language = f77define_macros = [('HAVE_CBLAS', None)]

3. 创建一个长度为10的空向量

np.zeros(10)

array([0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])

4.如何找到任何一个数组的内存大小

Z = np.zeros((10,10)) # 10*10为的矩阵
print(Z)
print("%d bytes" % (Z.size * Z.itemsize))

[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
800 bytes

5. 如何从命令行得到numpy中add函数的说明文档?

np.info(np.add) # 查看numpy工具包下的函数使用方法，很好用

add(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj])Add arguments element-wise.Parameters
----------
x1, x2 : array_likeThe arrays to be added. If ``x1.shape != x2.shape``, they must be broadcastable to a common shape (which becomes the shape of the output).
out : ndarray, None, or tuple of ndarray and None, optionalA location into which the result is stored. If provided, it must havea shape that the inputs broadcast to. If not provided or None,a freshly-allocated array is returned. A tuple (possible only as akeyword argument) must have length equal to the number of outputs.
where : array_like, optionalThis condition is broadcast over the input. At locations where thecondition is True, the `out` array will be set to the ufunc result.Elsewhere, the `out` array will retain its original value.Note that if an uninitialized `out` array is created via the default``out=None``, locations within it where the condition is False willremain uninitialized.
**kwargsFor other keyword-only arguments, see the:ref:`ufunc docs <ufuncs.kwargs>`.Returns
-------
add : ndarray or scalarThe sum of `x1` and `x2`, element-wise.This is a scalar if both `x1` and `x2` are scalars.Notes
-----
Equivalent to `x1` + `x2` in terms of array broadcasting.Examples
--------
>>> np.add(1.0, 4.0)
5.0
>>> x1 = np.arange(9.0).reshape((3, 3))
>>> x2 = np.arange(3.0)
>>> np.add(x1, x2)
array([[  0.,   2.,   4.],[  3.,   5.,   7.],[  6.,   8.,  10.]])

6. 创建一个长度为10并且除了第五个值为1的空向量

Z = np.zeros(10)
Z[4] = 11
print(Z)

[ 0.  0.  0.  0. 11.  0.  0.  0.  0.  0.]

7. 创建一个值域范围从10到49的向量

Z = np.arange(10,50)
Z

array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26,27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43,44, 45, 46, 47, 48, 49])

8. 反转一个向量(第一个元素变为最后一个)

Z = np.arange(50)
print(Z)
# 反转
Z = Z[ : :-1]
print(Z)

[ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 2324 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 4748 49]
[49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 2625 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10  9  8  7  6  5  4  3  21  0]

9.创建一个 3x3 并且值从0到8的矩阵

Z = np.arange(9).reshape(3,3)
Z

array([[0, 1, 2],[3, 4, 5],[6, 7, 8]])

10. 找到数组[1,2,0,0,4,0]中非0元素的位置索引

nz = np.nonzero([1,2,0,0,4,0])
nz

(array([0, 1, 4], dtype=int64),)

11. 创建一个 3x3 的单位矩阵

Z = np.eye(3)
Z

array([[1., 0., 0.],[0., 1., 0.],[0., 0., 1.]])

12. 创建一个 3x3x3的随机数组

Z = np.random.randint((3,3,3))
print(Z)
Z = np.random.random((3,3,3))
print(Z)

[0 1 0]
[[[0.11703768 0.15665001 0.98059487][0.6993327  0.12187568 0.3595888 ][0.00963102 0.2946496  0.21467848]][[0.96731389 0.98007911 0.18310304][0.09436081 0.52595679 0.14685079][0.38102119 0.28924822 0.59280225]][[0.02146097 0.79842838 0.91450769][0.65038213 0.78833875 0.85063171][0.24328237 0.53458248 0.67568819]]]

13. 创建一个 10x10 的随机数组并找到它的最大值和最小值

Z = np.random.random((10,10))
Zmin, Zmax = Z.min(), Z.max()
print(Zmin, Zmax)

0.010219243058963379 0.9890317300014234

14. 创建一个长度为30的随机向量并找到它的平均值

Z = np.random.random(30)
m = Z.mean()
print(m)

0.5125907170974495

15. 创建一个二维数组，其中边界值为1，其余值为0

Z = np.ones((10,10))
Z[1:-1,1:-1] = 0 # 这里怎么理解？
# 1到-1表示行列的位置，比如（1，1）（1,2）....(1,9),到第三行(2,1),(2,2)....(2,9),依次下去
print(Z)

[[1. 1. 1. 1. 1. 1. 1. 1. 1. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 0. 0. 0. 0. 0. 0. 0. 0. 1.][1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]]

16. 对于一个存在在数组，如何添加一个用0填充的边界?

Z = np.ones((5,5))
Z = np.pad(Z, pad_width=1, mode='constant', constant_values=0)
print(Z)

[[0. 0. 0. 0. 0. 0. 0.][0. 1. 1. 1. 1. 1. 0.][0. 1. 1. 1. 1. 1. 0.][0. 1. 1. 1. 1. 1. 0.][0. 1. 1. 1. 1. 1. 0.][0. 1. 1. 1. 1. 1. 0.][0. 0. 0. 0. 0. 0. 0.]]

看一看np.pad的用法
np.pad（）常用与深度学习中的数据预处理，可以将numpy数组按指定的方法填充成指定的形状。

方法参数：pad(array, pad_width, mode, **kwargs)
array：表示需要填充的数组；
pad_width：表示每个轴（axis）边缘需要填充的数值数目；
mode：表示填充的方式（取值：str字符串或用户提供的函数）

mode中的填充方式：
constant’——表示连续填充相同的值，每个轴可以分别指定填充值，constant_values=（x, y）
时前面用x填充，后面用y填充，缺省值填充0

17. 下面表达式运行的结果是什么？

 0 * np.nan

nan

np.nan == np.nan # 这里返回F，可以这样理解，无穷！=无穷

False

np.inf > np.nan

False

np.nan - np.nan

nan

0.3 == 3*0.1 #3*0.1=0.30000000000000004

False

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

Z = np.diag(1+np.arange(4),k=-1)
print(Z)

[[0 0 0 0 0][1 0 0 0 0][0 2 0 0 0][0 0 3 0 0][0 0 0 4 0]]

np.info(np.diag)

 diag(*args, **kwargs)Extract a diagonal or construct a diagonal array.See the more detailed documentation for ``numpy.diagonal`` if you use this
function to extract a diagonal and wish to write to the resulting array;
whether it returns a copy or a view depends on what version of numpy you
are using.Parameters
----------
v : array_likeIf `v` is a 2-D array, return a copy of its `k`-th diagonal.If `v` is a 1-D array, return a 2-D array with `v` on the `k`-thdiagonal.
k : int, optionalDiagonal in question. The default is 0. Use `k>0` for diagonalsabove the main diagonal, and `k<0` for diagonals below the maindiagonal.Returns
-------
out : ndarrayThe extracted diagonal or constructed diagonal array.See Also
--------
diagonal : Return specified diagonals.
diagflat : Create a 2-D array with the flattened input as a diagonal.
trace : Sum along diagonals.
triu : Upper triangle of an array.
tril : Lower triangle of an array.Examples
--------
>>> x = np.arange(9).reshape((3,3))
>>> x
array([[0, 1, 2],[3, 4, 5],[6, 7, 8]])>>> np.diag(x)
array([0, 4, 8])
>>> np.diag(x, k=1)
array([1, 5])
>>> np.diag(x, k=-1)
array([3, 7])>>> np.diag(np.diag(x))
array([[0, 0, 0],[0, 4, 0],[0, 0, 8]])

19. 创建一个8x8 的矩阵，并且设置成棋盘样式

Z = np.zeros((8,8),dtype=int)
Z[1::2,::2] = 1
Z[::2,1::2] = 1
print(Z)

[[0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0]]

#Z[i:j:s]
# 请参考https://blog.csdn.net/mingyuli/article/details/81604795的解释，我的理解是i和j大致是行列，s的话表示步长，当s<0时，Z[::-1]=a[-1:-len(a)-1:-1]

np.info(Z)

class:  ndarray
shape:  (5, 5)
strides:  (20, 4)
itemsize:  4
aligned:  True
contiguous:  True
fortran:  False
data pointer: 0x1b9eaedb1f0
byteorder:  little
byteswap:  False
type: int32

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?

print(np.unravel_index(100,(6,7,8)))

(1, 5, 4)

21. 用tile函数去创建一个 8x8的棋盘样式矩阵

Z = np.tile( np.array([[0,1],[1,0]]), (4,4))
print(Z)

[[0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0][0 1 0 1 0 1 0 1][1 0 1 0 1 0 1 0]]

np.tile()的用法 https://blog.csdn.net/qq_18433441/article/details/54897250

22. 对一个5x5的随机矩阵做归一化

Z = np.random.random((5,5))
Zmax, Zmin = Z.max(), Z.min()
Z = (Z - Zmin)/(Zmax - Zmin)
print(Z)

[[0.92042311 0.26179001 0.60885399 0.5080942  0.8458998 ][0.80656232 1.         0.52691876 0.64033586 0.23159865][0.71899013 0.61207932 0.01697865 0.55172094 0.28752424][0.99494883 0.02310809 0.71209696 0.80976307 0.        ][0.51834008 0.95287447 0.18104636 0.93275269 0.8524711 ]]

23. 创建一个将颜色描述为(RGBA)四个无符号字节的自定义dtype？

color = np.dtype([("r", np.ubyte, 1),("g", np.ubyte, 1),("b", np.ubyte, 1),("a", np.ubyte, 1)])
color

C:\Users\kingS\anaconda3\lib\site-packages\ipykernel_launcher.py:4: FutureWarning: Passing (type, 1) or '1type' as a synonym of type is deprecated; in a future version of numpy, it will be understood as (type, (1,)) / '(1,)type'.after removing the cwd from sys.path.dtype([('r', 'u1'), ('g', 'u1'), ('b', 'u1'), ('a', 'u1')])

24.一个5x3的矩阵与一个3x2的矩阵相乘，实矩阵乘积是什么？

Z = np.dot(np.ones((5,3)), np.ones((3,2)))
print(Z)

[[3. 3.][3. 3.][3. 3.][3. 3.][3. 3.]]

25. 给定一个一维数组，对其在3到8之间的所有元素取反

Z = np.arange(11)
Z[(3 < Z) & (Z <= 8)] *= -1
print(Z)

[ 0  1  2  3 -4 -5 -6 -7 -8  9 10]

26. 下面脚本运行后的结果是什么?

print(sum(range(5),-1))

print(sum(range(5),-1))

27. 考虑一个整数向量Z,下列表达合法的是哪个?

Z = np.arange(1,3)
Z

array([1, 2])

Z**Z

array([1, 4], dtype=int32)

2 << Z >> 2

array([1, 2], dtype=int32)

Z <- Z

array([False, False])

1j*Z

array([0.+1.j, 0.+2.j])

Z / 1/1

array([1., 2.])

Z < Z > Z

---------------------------------------------------------------------------ValueError                                Traceback (most recent call last)<ipython-input-87-a8a6f6561db3> in <module>
----> 1 Z < Z > ZValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

28. 下面表达式的结果分别是什么？

np.array(0) / np.array(0)

C:\Users\kingS\anaconda3\lib\site-packages\ipykernel_launcher.py:1: RuntimeWarning: invalid value encountered in true_divide"""Entry point for launching an IPython kernel.nan


np.array(0) // np.array(0)

C:\Users\kingS\anaconda3\lib\site-packages\ipykernel_launcher.py:1: RuntimeWarning: divide by zero encountered in floor_divide"""Entry point for launching an IPython kernel.0


np.array([np.nan]).astype(int).astype(float)        -2.14748365e+09

array([-4.2949673e+09])

29. 如何从零位开始舍入浮点数组？

Z = np.random.uniform(-10,+10,10)
print (np.copysign(np.ceil(np.abs(Z)), Z))

[ 1.  6.  9. -8.  5. -3. -2.  8. 10.  1.]

30. 如何找出两个数组公共的元素?

Z1 = np.random.randint(0, 10, 10)
Z2 = np.random.randint(0, 10, 10)
print (np.intersect1d(Z1, Z2))

[0 3 5 6 7]

np.random.uniform(-10,+10,10)

array([ 2.24073511, -4.68397267,  1.51454992, -8.12404611,  5.70720372,4.69355634,  5.93706946, -0.68425276,  4.40537982,  5.29810174])

np.ceil(np.random.uniform(-10,+10,10))

array([-5., -3.,  1.,  2., -8., -7.,  7., -6., -1., -0.])

31. 如何忽略所有的 numpy 警告(尽管不建议这么做)?

# Suicide mode on
defaults = np.seterr(all="ignore")
Z = np.ones(1) / 0# Back to sanity
_ = np.seterr(**defaults)# 另一个等价的方式， 使用上下文管理器（context manager）
with np.errstate(divide='ignore'):Z = np.ones(1) / 0
Z

array([inf])

32. 下面的表达式是否为真?

np.sqrt(-1) == np.emath.sqrt(-1)

C:\Users\kingS\anaconda3\lib\site-packages\ipykernel_launcher.py:1: RuntimeWarning: invalid value encountered in sqrt"""Entry point for launching an IPython kernel.False

33. 如何获得昨天，今天和明天的日期?

yesterday = np.datetime64('today', 'D') - np.timedelta64(1, 'D')
today = np.datetime64('today', 'D')
tomorrow = np.datetime64('today', 'D') + np.timedelta64(1, 'D')
tomorrow

numpy.datetime64('2021-06-04')

np.datetime64('today', 'D')

numpy.datetime64('2021-06-03')

34. 怎么获得所有与2016年7月的所有日期?

Z = np.arange('2016-07', '2016-08', dtype='datetime64[D]')
print (Z)

['2016-07-01' '2016-07-02' '2016-07-03' '2016-07-04' '2016-07-05''2016-07-06' '2016-07-07' '2016-07-08' '2016-07-09' '2016-07-10''2016-07-11' '2016-07-12' '2016-07-13' '2016-07-14' '2016-07-15''2016-07-16' '2016-07-17' '2016-07-18' '2016-07-19' '2016-07-20''2016-07-21' '2016-07-22' '2016-07-23' '2016-07-24' '2016-07-25''2016-07-26' '2016-07-27' '2016-07-28' '2016-07-29' '2016-07-30''2016-07-31']

35. 如何计算 ((A+B)*(-A/2)) (不使用中间变量)?

A = np.ones(3) * 1
B = np.ones(3) * 1
C = np.ones(3) * 1
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., -1., -1.])

36. 用5种不同的方法提取随机数组中的整数部分

Z = np.random.uniform(0, 10, 10)
print(Z)
print (Z - Z % 1) #
print (np.floor(Z))
print (np.ceil(Z)-1)
print (Z.astype(int))
print (np.trunc(Z))

[6.80650555 1.74342436 2.70167401 0.73914672 0.94154516 7.279028738.63727446 8.81343166 8.8309756  7.5328844 ]
[6. 1. 2. 0. 0. 7. 8. 8. 8. 7.]
[6. 1. 2. 0. 0. 7. 8. 8. 8. 7.]
[6. 1. 2. 0. 0. 7. 8. 8. 8. 7.]
[6 1 2 0 0 7 8 8 8 7]
[6. 1. 2. 0. 0. 7. 8. 8. 8. 7.]

1.2 % 1

0.19999999999999996

37. 创建一个5x5的矩阵且每一行的值范围为从0到4

Z = np.zeros((5, 5))
Z += np.arange(5)
print (Z)

[[0. 1. 2. 3. 4.][0. 1. 2. 3. 4.][0. 1. 2. 3. 4.][0. 1. 2. 3. 4.][0. 1. 2. 3. 4.]]

38. 如何用一个生成10个整数的函数来构建数组

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.]

np.fromiter用法和解释 https://blog.csdn.net/MaeveShi/article/details/107388473

39. 创建一个大小为10的向量，值域为0到1，不包括0和1

Z = np.linspace(0, 1, 12, endpoint=True)[1: -1]
print (Z)

[0.09090909 0.18181818 0.27272727 0.36363636 0.45454545 0.545454550.63636364 0.72727273 0.81818182 0.90909091]

np.linspace(0, 1, 12, endpoint=True)

array([0.        , 0.08333333, 0.16666667, 0.25      , 0.33333333,0.41666667, 0.5       , 0.58333333, 0.66666667, 0.75      ,0.83333333, 0.91666667])

np.info(np.linspace)

 linspace(*args, **kwargs)Return evenly spaced numbers over a specified interval.Returns `num` evenly spaced samples, calculated over the
interval [`start`, `stop`].The endpoint of the interval can optionally be excluded... versionchanged:: 1.16.0Non-scalar `start` and `stop` are now supported.Parameters
----------
start : array_likeThe starting value of the sequence.
stop : array_likeThe end value of the sequence, unless `endpoint` is set to False.In that case, the sequence consists of all but the last of ``num + 1``evenly spaced samples, so that `stop` is excluded.  Note that the stepsize changes when `endpoint` is False.
num : int, optionalNumber of samples to generate. Default is 50. Must be non-negative.
endpoint : bool, optionalIf True, `stop` is the last sample. Otherwise, it is not included.Default is True.
retstep : bool, optionalIf True, return (`samples`, `step`), where `step` is the spacingbetween samples.
dtype : dtype, optionalThe type of the output array.  If `dtype` is not given, infer the datatype from the other input arguments... versionadded:: 1.9.0axis : int, optionalThe axis in the result to store the samples.  Relevant only if startor stop are array-like.  By default (0), the samples will be along anew axis inserted at the beginning. Use -1 to get an axis at the end... versionadded:: 1.16.0Returns
-------
samples : ndarrayThere are `num` equally spaced samples in the closed interval``[start, stop]`` or the half-open interval ``[start, stop)``(depending on whether `endpoint` is True or False).
step : float, optionalOnly returned if `retstep` is TrueSize of spacing between samples.See Also
--------
arange : Similar to `linspace`, but uses a step size (instead of thenumber of samples).
geomspace : Similar to `linspace`, but with numbers spaced evenly on a logscale (a geometric progression).
logspace : Similar to `geomspace`, but with the end points specified aslogarithms.Examples
--------
>>> np.linspace(2.0, 3.0, num=5)
array([2.  , 2.25, 2.5 , 2.75, 3.  ])
>>> np.linspace(2.0, 3.0, num=5, endpoint=False)
array([2. ,  2.2,  2.4,  2.6,  2.8])
>>> np.linspace(2.0, 3.0, num=5, retstep=True)
(array([2.  ,  2.25,  2.5 ,  2.75,  3.  ]), 0.25)Graphical illustration:>>> import matplotlib.pyplot as plt
>>> N = 8
>>> y = np.zeros(N)
>>> x1 = np.linspace(0, 10, N, endpoint=True)
>>> x2 = np.linspace(0, 10, N, endpoint=False)
>>> plt.plot(x1, y, 'o')
[<matplotlib.lines.Line2D object at 0x...>]
>>> plt.plot(x2, y + 0.5, 'o')
[<matplotlib.lines.Line2D object at 0x...>]
>>> plt.ylim([-0.5, 1])
(-0.5, 1)
>>> plt.show()

40. 创建一个大小为10的随机向量，并把它排序

Z = np.random.random(10)
Z.sort()
print (Z)

[0.10705316 0.11344548 0.45293034 0.45776226 0.5286275  0.542591170.56924058 0.69027986 0.7160883  0.7170026 ]

41. 对一个小数组进行求和有没有办法比np.sum更快?

Z = np.arange(10)
np.add.reduce(Z) # rdcuce感觉像是累（加、减、乘）

np.info(np.add.reduce)

reduce(a, axis=0, dtype=None, out=None, keepdims=False, initial=<no value>, where=True)Reduces `a`'s dimension by one, by applying ufunc along one axis.Let :math:`a.shape = (N_0, ..., N_i, ..., N_{M-1})`.  Then
:math:`ufunc.reduce(a, axis=i)[k_0, ..,k_{i-1}, k_{i+1}, .., k_{M-1}]` =
the result of iterating `j` over :math:`range(N_i)`, cumulatively applying
ufunc to each :math:`a[k_0, ..,k_{i-1}, j, k_{i+1}, .., k_{M-1}]`.
For a one-dimensional array, reduce produces results equivalent to:
::r = op.identity # op = ufuncfor i in range(len(A)):r = op(r, A[i])return rFor example, add.reduce() is equivalent to sum().Parameters
----------
a : array_likeThe array to act on.
axis : None or int or tuple of ints, optionalAxis or axes along which a reduction is performed.The default (`axis` = 0) is perform a reduction over the firstdimension of the input array. `axis` may be negative, inwhich case it counts from the last to the first axis... versionadded:: 1.7.0If this is None, a reduction is performed over all the axes.If this is a tuple of ints, a reduction is performed on multipleaxes, instead of a single axis or all the axes as before.For operations which are either not commutative or not associative,doing a reduction over multiple axes is not well-defined. Theufuncs do not currently raise an exception in this case, but willlikely do so in the future.
dtype : data-type code, optionalThe type used to represent the intermediate results. Defaultsto the data-type of the output array if this is provided, orthe data-type of the input array if no output array is provided.
out : ndarray, None, or tuple of ndarray and None, optionalA location into which the result is stored. If not provided or None,a freshly-allocated array is returned. For consistency with``ufunc.__call__``, if given as a keyword, this may be wrapped in a1-element tuple... versionchanged:: 1.13.0Tuples are allowed for keyword argument.
keepdims : bool, optionalIf this is set to True, the axes which are reduced are leftin the result as dimensions with size one. With this option,the result will broadcast correctly against the original `arr`... versionadded:: 1.7.0
initial : scalar, optionalThe value with which to start the reduction.If the ufunc has no identity or the dtype is object, this defaultsto None - otherwise it defaults to ufunc.identity.If ``None`` is given, the first element of the reduction is used,and an error is thrown if the reduction is empty... versionadded:: 1.15.0where : array_like of bool, optionalA boolean array which is broadcasted to match the dimensionsof `a`, and selects elements to include in the reduction. Notethat for ufuncs like ``minimum`` that do not have an identitydefined, one has to pass in also ``initial``... versionadded:: 1.17.0Returns
-------
r : ndarrayThe reduced array. If `out` was supplied, `r` is a reference to it.Examples
--------
>>> np.multiply.reduce([2,3,5])
30A multi-dimensional array example:>>> X = np.arange(8).reshape((2,2,2))
>>> X
array([[[0, 1],[2, 3]],[[4, 5],[6, 7]]])
>>> np.add.reduce(X, 0)
array([[ 4,  6],[ 8, 10]])
>>> np.add.reduce(X) # confirm: default axis value is 0
array([[ 4,  6],[ 8, 10]])
>>> np.add.reduce(X, 1)
array([[ 2,  4],[10, 12]])
>>> np.add.reduce(X, 2)
array([[ 1,  5],[ 9, 13]])You can use the ``initial`` keyword argument to initialize the reduction
with a different value, and ``where`` to select specific elements to include:>>> np.add.reduce([10], initial=5)
15
>>> np.add.reduce(np.ones((2, 2, 2)), axis=(0, 2), initial=10)
array([14., 14.])
>>> a = np.array([10., np.nan, 10])
>>> np.add.reduce(a, where=~np.isnan(a))
20.0Allows reductions of empty arrays where they would normally fail, i.e.
for ufuncs without an identity.>>> np.minimum.reduce([], initial=np.inf)
inf
>>> np.minimum.reduce([[1., 2.], [3., 4.]], initial=10., where=[True, False])
array([ 1., 10.])
>>> np.minimum.reduce([])
Traceback (most recent call last):...
ValueError: zero-size array to reduction operation minimum which has no identity

42. 如何判断两和随机数组相等

A = np.random.randint(0, 2, 5)
B = np.random.randint(0, 2, 5)# 假设array的形状(shape)相同和一个误差容限（tolerance）
equal = np.allclose(A,B)
print(equal)# 检查形状和元素值，没有误差容限（值必须完全相等）
equal = np.array_equal(A,B)
print(equal)

False
False

43. 把数组变为只读

Z = np.zeros(5)
Z.flags.writeable = True # 改成False会报错
Z[0] = 1
Z

array([1., 0., 0., 0., 0.])

44. 将一个10x2的笛卡尔坐标矩阵转换为极坐标

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)

[0.99476243 0.18120069 0.24878277 0.35681154 1.02005226 0.796211161.06348455 1.25476229 0.75381007 0.76924139]
[0.83628677 1.31222218 1.18569439 0.28174263 0.36789134 0.08009951.16991472 0.71520923 0.83252375 1.04955151]

np.random.random((10, 2))

array([[0.54801274, 0.22043258],[0.14532328, 0.96551325],[0.24424964, 0.99437072],[0.79961282, 0.94123918],[0.43329684, 0.79543283],[0.73901185, 0.44119212],[0.6425194 , 0.47897277],[0.66040064, 0.41569665],[0.92180864, 0.78082399],[0.72624301, 0.21686495]])

45. 创建一个大小为10的随机向量并且将该向量中最大的值替换为0

Z = np.random.random(10)
Z[Z.argmax()] = 0
print (Z)

[0.0027154  0.11186707 0.73458462 0.85812928 0.         0.130172520.68314185 0.32608012 0.90231489 0.39966661]

Z = np.random.random(10)

Z.argmax()# 返回第几个最大

Z.max()

0.9270562760783981

46. 创建一个结构化数组，其中x和y坐标覆盖[0, 1]x[1, 0]区域

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)

[[(0.  , 0.  ) (0.25, 0.  ) (0.5 , 0.  ) (0.75, 0.  ) (1.  , 0.  )][(0.  , 0.25) (0.25, 0.25) (0.5 , 0.25) (0.75, 0.25) (1.  , 0.25)][(0.  , 0.5 ) (0.25, 0.5 ) (0.5 , 0.5 ) (0.75, 0.5 ) (1.  , 0.5 )][(0.  , 0.75) (0.25, 0.75) (0.5 , 0.75) (0.75, 0.75) (1.  , 0.75)][(0.  , 1.  ) (0.25, 1.  ) (0.5 , 1.  ) (0.75, 1.  ) (1.  , 1.  )]]

47. 给定两个数组X和Y，构造柯西(Cauchy)矩阵C

[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-YFILBw7c-1622651660496)(attachment:image.png)]

X = np.arange(8)
Y = X + 0.5
C = 1.0 / np.subtract.outer(X, Y)
print (C)
print(np.linalg.det(C)) # 计算行列式

[[-2.         -0.66666667 -0.4        -0.28571429 -0.22222222 -0.18181818-0.15384615 -0.13333333][ 2.         -2.         -0.66666667 -0.4        -0.28571429 -0.22222222-0.18181818 -0.15384615][ 0.66666667  2.         -2.         -0.66666667 -0.4        -0.28571429-0.22222222 -0.18181818][ 0.4         0.66666667  2.         -2.         -0.66666667 -0.4-0.28571429 -0.22222222][ 0.28571429  0.4         0.66666667  2.         -2.         -0.66666667-0.4        -0.28571429][ 0.22222222  0.28571429  0.4         0.66666667  2.         -2.-0.66666667 -0.4       ][ 0.18181818  0.22222222  0.28571429  0.4         0.66666667  2.-2.         -0.66666667][ 0.15384615  0.18181818  0.22222222  0.28571429  0.4         0.666666672.         -2.        ]]
3638.163637117973

# 这点好像是没看懂，涉及到数学了

48. 打印每个numpy 类型的最小和最大可表示值

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)

-128
127
-2147483648
2147483647
-9223372036854775808
9223372036854775807
-3.4028235e+38
3.4028235e+38
1.1920929e-07
-1.7976931348623157e+308
1.7976931348623157e+308
2.220446049250313e-16

49. 如何打印数组中所有的值？

np.set_printoptions(threshold=10000)# np.nan替换10000会报错
Z = np.zeros((16,16))
print(Z)

[[0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.][0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]

50. 如何在数组中找到与给定标量接近的值?

Z = np.arange(100)
v = np.random.uniform(0, 100)
index = (np.abs(Z-v)).argmin()
print(Z[index])
# 结果：18

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