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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?
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?
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]]
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。
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]
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]
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]
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
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:
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:
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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