The default, axis=None, will sum all of the elements of the input array. If axis is negative it counts from the last to the first axis. If axis is a tuple of ints, a sum is performed on all of the axes specified in the tuple instead of a single axis or all the axes as before...
find_objectsdata = numpy.array([0, 0, 0, 2, 2, 0, 2, 2, 2, 0])labels, number_of_regions = label(a)ranges = find_objects(labels)for identified_range in ranges: print identified_range[0].start, identified_range[0].stop您应该看到:3 56 9希望这可以帮助!
a=np.array([10,20,30,40]) # array([10, 20, 30, 40]) b=np.arange(4) # array([0, 1, 2, 3]) numpy 的几种基本运算 ¶ 上述代码中的 a 和 b 是两个属性为 array 也就是矩阵的变量,而且二者都是1行4列的矩阵, 其中b矩阵中的元素分别是从0到3。 如果我们想要求两个矩阵之间的减法...
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 报错原因: Numpy对逻辑表达式判别不清楚,它可以返回False如果等号两边两个式子是数值相等,也可以返回True因为等号两边两个式子是逻辑相等。...
>>> np.array_equal(a, c) True 逻辑运算: >>> >>> a = np.array([1, 1, 0, 0], dtype=bool) >>> b = np.array([1, 0, 1, 0], dtype=bool) >>> np.logical_or(a, b) array([ True, True, True, False]) >>> np.logical_and(a, b) ...
numpy.array:创建新的NumPy数组 # Create an array using np.array() arr = np.array([1, 2, 3, 4, 5]) print(arr) Ouput: [1 2 3 4 5] numpy.zeros:创建一个以零填充的数组。 # Create a 2-dimensional array of zeros arr = np.zeros((3, 4)) ...
import numpy as np # 创建不同类型的数组 lengths = np.array([2.5, 3.8, 4.1], dtype=np.float32) times = np.array([20, 35, 55], dtype=np.int64) flags = np.array([True, False, True], dtype=np.bool_) # 检查数组的数据类型 print("Data Types:", lengths.dtype, times.dtype, flags...
print('the result of a+b :',d) print('the result of b**2 :',e) print('the result of sin(a) :',np.sin(a)) a <45# 返回布尔值 array([ True, True, True, False]) # 矩阵运算 A = np.array( [[1,1] ,[0,1]] ) ...
b = np.array([2,4,6]) # Stack two arrays row-wise print(np.vstack((a,b))) >>>[[135] [246]] # Stack two arrays column-wise print(np.hstack((a,b))) >>>[135246] 分割数组 举例: # Split array into groups of ~3
# > [ True, True, True]], dtype=bool) # Alternate method: np.ones((3,3), dtype=bool) 3、将一维数组转换为2行的2维数组 arr = np.arange(10) arr.reshape(2, -1) # Setting to -1 automatically decides the number of cols # > array([[0, 1, 2, 3, 4], ...