task 05: 排序,搜索,计数和集合

排序

numpy.sort(a[, axis=-1, kind=‘quicksort’, order=None]) Return a sorted copy of an array.
a. axis:排序沿数组的(轴)方向,0表示按行,1表示按列,None表示展开来排序,默认为-1,表示沿最后的轴排序。
b. kind:排序的算法,提供了快排’quicksort’、混排’mergesort’、堆排’heapsort’, 默认为‘quicksort’。
c. order:排序的字段名,可指定字段排序,默认为None。

import numpy as np
np.random.seed(20200612)
x = np.random.rand(5, 5) * 10
x = np.around(x, 2)
print(x)
# [[2.32 7.54 9.78 1.73 6.22]
# [6.93 5.17 9.28 9.76 8.25]
# [0.01 4.23 0.19 1.73 9.27]
# [7.99 4.97 0.88 7.32 4.29]
# [9.05 0.07 8.95 7.9 6.99]]
y = np.sort(x)
print(y)
# [[1.73 2.32 6.22 7.54 9.78]
# [5.17 6.93 8.25 9.28 9.76]
# [0.01 0.19 1.73 4.23 9.27]
# [0.88 4.29 4.97 7.32 7.99]
# [0.07 6.99 7.9 8.95 9.05]]
y = np.sort(x, axis=0)
print(y)
# [[0.01 0.07 0.19 1.73 4.29]
# [2.32 4.23 0.88 1.73 6.22]
# [6.93 4.97 8.95 7.32 6.99]
# [7.99 5.17 9.28 7.9 8.25]
# [9.05 7.54 9.78 9.76 9.27]]
y = np.sort(x, axis=1)
print(y)
# [[1.73 2.32 6.22 7.54 9.78]
# [5.17 6.93 8.25 9.28 9.76]
# [0.01 0.19 1.73 4.23 9.27]
# [0.88 4.29 4.97 7.32 7.99]
# [0.07 6.99 35 7.9 8.95 9.05]]
import numpy as np
dt = np.dtype([('name', 'S10'), ('age', np.int)])
#将包含多个字段的结构化数据应用于ndarray对象
a = np.array([("Mike", 21), ("Nancy", 25), ("Bob", 17), ("Jane", 27)], dtype=dt)
b = np.sort(a, order='name')
print(b)
# [(b'Bob', 17) (b'Jane', 27) (b'Mike', 21) (b'Nancy', 25)]
b = np.sort(a, order='age')
print(b)
# [(b'Bob', 17) (b'Mike', 21) (b'Nancy', 25) (b'Jane', 27)]

如果排序后,想用元素的索引位置替代排序后的实际结果,该怎么办呢?

  1. numpy.argsort(a[, axis=-1, kind=‘quicksort’, order=None]) Returns the indices that would sort an array.

对数组沿给定轴执行间接排序,并使用指定排序类型返回数据的索引数组。这个索引数组用于构造排序后的数组。

import numpy as np
np.random.seed(20200612)
x = np.random.randint(0, 10, 10)
print(x)
# [6 1 8 5 5 4 1 2 9 1]
y = np.argsort(x)
print(y)
# [1 6 9 7 5 3 4 0 2 8]   返回从小到大的索引
print(x[y])
# [1 1 1 2 4 5 5 6 8 9]
y = np.argsort(-x)
print(y)
# [8 2 0 3 4 5 7 1 6 9]
print(x[y])
# [9 8 6 5 5 4 2 1 1 1]

如何将数据按照某一指标进行排序呢?

  1. numpy.lexsort(keys[, axis=-1]) Perform an indirect stable sort using a sequence of keys.(使用键序列执行间接稳定排序。)
  2. 给定多个可以在电子表格中解释为列的排序键,lexsort返回一个整数索引数组,该数组描述了按多个列排序的顺序。序列中的最后一个键用于主排序顺序,倒数第二个键用于辅助排序顺序,依此类推。keys参数必须是可以转换为相同形状的数组的对象序列。如果为keys参数提供了2D数组,则将其行解释为排序键,并根据最后一行,倒数第二行等进行排序。
import numpy as np
np.random.seed(20200612)
x = np.random.rand(5, 5) * 10
x = np.around(x, 2)
print(x)
# [[2.32 7.54 9.78 1.73 6.22]
# [6.93 5.17 9.28 9.76 8.25]
# [0.01 4.23 0.19 1.73 9.27]
# [7.99 4.97 0.88 7.32 4.29]
# [9.05 0.07 8.95 7.9 6.99]]
index = np.lexsort([x[:, 0]])
print(index)
# [2 0 1 3 4]
y = x[index]
print(y)
# [[0.01 4.23 0.19 1.73 9.27]
# [2.32 7.54 9.78 1.73 6.22]
# [6.93 5.17 9.28 9.76 8.25]
# [7.99 4.97 0.88 7.32 4.29]
# [9.05 0.07 8.95 7.9 6.99]]

参考lexsort

  1. numpy.partition(a, kth, axis=-1, kind=‘introselect’, order=None) Return a partitioned copy of an array.
    Creates a copy of the array with its elements rearranged in such a way that the value of the element in k-th position is in the position it would be in a sorted array. All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. The ordering of the elements in the two partitions is undefined.
    【例】以索引是 kth 的元素为基准,将元素分成两部分,即大于该元素的放在其后面,小于该元素的放在其前面,这里有点类似于快排。

参看partition

  1. numpy.argpartition(a, kth, axis=-1, kind=‘introselect’, order=None)
    Perform an indirect partition along the given axis using the algorithm specified by the kind keyword. It returns an array of indices of the same shape as a that index data along the given axis in partitioned order.

搜索

numpy.argmax(a[, axis=None, out=None])
Returns the indices of the maximum values along an axis.
numpy.argmin(a[, axis=None, out=None])
Returns the indices of the minimum values along an axis.

import numpy as np
np.random.seed(20200612)
x = np.random.rand(5, 5) * 10
x = np.around(x, 2)
print(x)
# [[2.32 7.54 9.78 1.73 6.22]
# [6.93 5.17 9.28 9.76 8.25]
# [0.01 4.23 0.19 1.73 9.27]
# [7.99 4.97 0.88 7.32 4.29]
# [9.05 0.07 8.95 7.9 6.99]]
y = np.argmax(x)
print(y) # 2  看作是一维的数列,然后给出索引
y = np.argmax(x, axis=0)
print(y)
# [4 0 0 1 2]
y = np.argmax(x, axis=1)
print(y)
# [2 3 4 0 0]

numppy.nonzero(a) Return the indices of the elements that are non-zero.
,其值为非零元素的下标在对应轴上的值。

  1. 只有a 中非零元素才会有索引值,那些零值元素没有索引值。
  2. 返回一个长度为a.ndim 的元组(tuple),元组的每个元素都是一个整数数组(array)。
  3. 每一个array均是从一个维度上来描述其索引值。比如,如果a 是一个二维数组,则tuple包含两个array,第一个array从行维度来描述索引
    值;第二个array从列维度来描述索引值。
  4. 该 np.transpose(np.nonzero(x)) 函数能够描述出每一个非零元素在不同维度的索引值。
  5. 通过a[nonzero(a)] 得到所有a 中的非零值。
import numpy as np
x = np.array([0, 2, 3])
print(x) # [0 2 3]
print(x.shape) # (3,)
print(x.ndim) # 1
y = np.nonzero(x)
print(y) # (array([1, 2], dtype=int64),)
print(np.array(y)) # [[1 2]]
print(np.array(y).shape) # (1, 2)
print(np.array(y).ndim) # 2
print(np.transpose(y))
# [[1]
# [2]]
print(x[np.nonzero(x)])
#[2, 3]

参考nonzero

nonzero() 将布尔数组转换成整数数组进行操作.

import numpy as np
x = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
print(x)
# [[1 2 3]
# [4 5 6]
# [7 8 9]]
y = x > 3
print(y)
# [[False False False]
# [ True True True]
# [ True True True]]
y = np.nonzero(x > 3)
print(y)
# (array([1, 1, 1, 2, 2, 2], dtype=int64), array([0, 1, 2, 0, 1, 2], dtype=int64))
y = x[np.nonzero(x > 3)]
print(y)
# [4 5 6 7 8 9]
y = x[x > 3]
print(y)
# [4 5 6 7 8 9]
  1. numpy.where(condition, [x=None, y=None]) Return elements chosen from x or y depending on condition .
    【例】满足条件condition ,输出x ,不满足输出y 。
import numpy as np
x = np.arange(10)
print(x)
# [0 1 2 3 4 5 6 7 8 9]
y = np.where(x < 5, x, 10 * x)
print(y)
# [ 0 1 2 3 4 50 60 70 80 90]
x = np.array([[0, 1, 2],
[0, 2, 4],
[0, 3, 6]])
y = np.where(x < 4, x, -1)
print(y)
# [[ 0 1 2]
# [ 0 2 -1]
# [ 0 3 -1]]

只有condition ,没有x 和y ,则输出满足条件 (即非0) 元素的坐标 (等价于numpy.nonzero )。这里的坐标以tuple的形式给出,通常原数组有多少维,输出的tuple中就包含几个数组,分别对应符合条件元素的各维坐标。

x = np.array([[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]])
y = np.where(x > 25)
print(y)
# (array([3, 3, 3, 3, 3, 4, 4, 4, 4, 4], dtype=int64), array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4], dtype=int64))
print(x[y])
# [26 27 28 29 30 31 32 33 34 35]
y = np.nonzero(x > 25)
print(y)
# (array([3, 3, 3, 3, 3, 4, 4, 4, 4, 4], dtype=int64), array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4], dtype=int64))
print(x[y])
# [26 27 28 29 30 31 32 33 34 35]

numpy.searchsorted(a, v[, side=‘left’, sorter=None]) Find indices where elements should be inserted to maintain order.
a. a:一维输入数组。当sorter 参数为None 的时候, a 必须为升序数组;否则, sorter 不能为空,存放a 中元素的index ,用于反映a 数组的升序排列方式。
b. v:插入a 数组的值,可以为单个元素, list 或者ndarray 。
c. side:查询方向,当为left 时,将返回第一个符合条件的元素下标;当为right 时,将返回最后一个符合条件的元素下标。
d. sorter:一维数组存放a 数组元素的 index,index 对应元素为升序。

import numpy as np
x = np.array([0, 1, 5, 9, 11, 18, 26, 33])
y = np.searchsorted(x, 15)
print(y) # 5
y = np.searchsorted(x, 15, side='right')
print(y) # 5
y = np.searchsorted(x, 11)
print(y) # 4
y = np.searchsorted(x, 11, side='right')
print(y) # 5
import numpy as np
x = np.array([0, 1, 5, 9, 11, 18, 26, 33])
np.random.shuffle(x)    #打乱顺序
print(x) # [33 1 9 18 11 26 0 5]
x_sort = np.argsort(x)
print(x_sort) # [6 1 7 2 4 3 5 0]
y = np.searchsorted(x, [-1, 0, 11, 15, 33, 35], sorter=x_sort)
print(y) # [0 0 4 5 7 8]
y = np.searchsorted(x, [-1, 0, 11, 15, 33, 35], side='right', sorter=x_sort)
print(y) # [0 1 5 5 8 8]

计数

numpy.count_nonzero(a, axis=None) Counts the number of non-zero values in the array a.

import numpy as np
x = np.count_nonzero(np.eye(4))
print(x) # 4
x = np.count_nonzero([[0, 1, 7, 0, 0], [3, 0, 0, 2, 19]])
print(x) # 5
x = np.count_nonzero([[0, 1, 7, 0, 0], 
                      [3, 0, 0, 2, 19]], axis=0)
print(x) # [1 1 1 1 1]
x = np.count_nonzero([[0, 1, 7, 0, 0], 
                      [3, 0, 0, 2, 19]], axis=1)
print(x) # [2 3] #第一行两个非零,第二行三个非零

集合操作

构造集合
numpy.unique(ar, return_index=False, return_inverse=False, return_counts=False, axis=None) Find the unique elements of an array.
a. return_index=True 表示返回新列表元素在旧列表中的位置。
b. return_inverse=True 表示返回旧列表元素在新列表中的位置。
c. return_counts=True 表示返回新列表元素在旧列表中出现的次数。

import numpy as np
x = np.unique([1, 1, 3, 2, 3, 3])
print(x) # [1 2 3]
x = sorted(set([1, 1, 3, 2, 3, 3]))
print(x) # [1, 2, 3]
x = np.array([[1, 1], [2, 3]])
u = np.unique(x)
print(u) # [1 2 3]
x = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])
y = np.unique(x, axis=0)
print(y)
# [[1 0 0]
# [2 3 4]]
x = np.array(['a', 'b', 'b', 'c', 'a'])
u, index = np.unique(x, return_index=True)
print(u) # ['a' 'b' 'c']
print(index) # [0 1 3]
print(x[index]) # ['a' 'b' 'c']
x = np.array([1, 2, 6, 4, 2, 3, 2])
u, index = np.unique(x, return_inverse=True)
print(u) # [1 2 3 4 6]
print(index) # [0 1 4 3 1 2 1]
print(u[index]) # [1 2 6 4 2 3 2]
u, count = np.unique(x, return_counts=True)
print(u) # [1 2 3 4 6]
print(count) # [1 3 1 1 1]

布尔运算

numpy.in1d(ar1, ar2, assume_unique=False, invert=False) Test whether each element of a 1-D array is also present in a second array.
Returns a boolean array the same length as ar1 that is True where an element of ar1 is in ar2 and False otherwise.
【例】前面的数组是否包含于后面的数组,返回布尔值。返回的值是针对第一个参数的数组的,所以维数和第一个参数一致,布尔值与数组的元素位置也一一对应。

import numpy as np
test = np.array([0, 1, 2, 5, 0])
states = [0, 2]
mask = np.in1d(test, states)
print(mask) # [ True False True False True]
print(test[mask]) # [0 2 0]
mask = np.in1d(test, states, invert=True)
print(mask) # [False True False True False]
print(test[mask]) # [1 5]

求两个集合的交集
numpy.intersect1d(ar1, ar2, assume_unique=False, return_indices=False) Find the intersection of two arrays.
Return the sorted, unique values that are in both of the input arrays.
【例】求两个数组的唯一化+求交集+排序函数。

import numpy as np
from functools import reduce
x = np.intersect1d([1, 3, 4, 3], [3, 1, 2, 1])
print(x) # [1 3]
x = np.array([1, 1, 2, 3, 4])
y = np.array([2, 1, 4, 6])
xy, x_ind, y_ind = np.intersect1d(x, y, return_indices=True)
print(x_ind) # [0 2 4]
print(y_ind) # [1 0 2]
print(xy) # [1 2 4]
print(x[x_ind]) # [1 2 4]
print(y[y_ind]) # [1 2 4]
x = reduce(np.intersect1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
print(x) # [3]

求两个集合的并集:
numpy.union1d(ar1, ar2) Find the union of two arrays.
Return the unique, sorted array of values that are in either of the two input arrays.

import numpy as np
from functools import reduce
x = np.union1d([-1, 0, 1], [-2, 0, 2])
print(x) # [-2 -1 0 1 2]
x = reduce(np.union1d, ([1, 3, 4, 3], [3, 1, 2, 1], [6, 3, 4, 2]))
print(x) # [1 2 3 4 6]
'''
functools.reduce(function, iterable[, initializer])
将两个参数的 function 从左至右积累地应用到 iterable 的条目,
以便将该可迭代对象缩减为单一的值。 例如,reduce(lambda x, y:x+y, [1, 2, 3, 4, 5]) 
是计算 ((((1+2)+3)+4)+5) 的值。 左边的参数 x 是积累值而右边的参数 y 则是来自 iterable 的更新值。 如果存在可选项 initializer,它会被放在参与计算的可迭代对象的条目之前,并在可迭代对象为空时作为默认值。 如果没有给出 initializer并且 iterable 仅包含一个条目,则将返回第一项。
大致相当于:
def reduce(function, iterable, initializer=None):
it = iter(iterable)
if initializer is None:
value = next(it)
else:
value = initializer
for element in it:
value = function(value, element)
return value
'''

求两个集合的差集:

  1. numpy.setdiff1d(ar1, ar2, assume_unique=False) Find the set difference of two arrays.
    Return the unique values in ar1 that are not in ar2 .
    【例】集合的差,即元素存在于第一个函数不存在于第二个函数中。
import numpy as np
a = np.array([1, 2, 3, 2, 4, 1])
b = np.array([3, 4, 5, 6])
x = np.setdiff1d(a, b)
print(x) # [1 2]

求两个集合的异或:

  1. setxor1d(ar1, ar2, assume_unique=False) Find the set exclusive-or of two arrays.
    【例】集合的对称差,即两个集合的交集的补集。简言之,就是两个数组中各自独自拥有的元素的集合。
import numpy as np
a = np.array([1, 2, 3, 2, 4, 1])
b = np.array([3, 4, 5, 6])
x = np.setxor1d(a, b)
print(x) # [1 2 5 6]

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