a = np.random.random([100,3]) b = np.arange(100).reshape([100,1]) a = np.concatenate([a,b],axis=-1) c = np.random.random([7,3]) d = np.array([1,5,8,11,52,45,31]) c = np.concatenate([c,d],axis=-1) 按照c中的-1列代表的位置更新a的所有数值

可以通过以下代码实现：

import numpy as np

a = np.random.random([100, 3])
b = np.arange(100).reshape([100, 1])
a = np.concatenate([a, b], axis=-1)

c = np.random.random([7, 3])
d = np.array([1, 5, 8, 11, 52, 45, 31])
c = np.concatenate([c, d.reshape(-1, 1)], axis=-1)

a[:, -1] = c[:, -1]

print(a)

这里先将`d`通过`reshape`函数转换成列向量，然后将`c`和`d`沿着列方向合并，得到更新后的`c`。然后，直接将`c`的最后一列赋值给`a`的最后一列即可。最终输出更新后的`a`。

相关问题

arr_a = np.arange(8).reshape(2, 4) arr_b= np.random.randint(1, 10, size = 4),通过python将arr_a和arr_b转换成矩阵，计算矩阵的矢量积和数量积。

import numpy as np

arr_a = np.arange(8).reshape(2, 4)
arr_b = np.random.randint(1, 10, size=4)

# 转换为矩阵
mat_a = np.matrix(arr_a)
mat_b = np.matrix(arr_b)

# 计算矢量积
vec_product = np.dot(mat_a, mat_b.T)
print("矢量积：")
print(vec_product)

# 计算数量积
num_product = np.dot(mat_a.flatten(), mat_b.T.flatten())
print("数量积：")
print(num_product)

import numpy as np import pandas as pd def localmin(points, pixel_size): x_min = np.min(points[:, 0]) y_min = np.min(points[:, 1]) x_max = np.max(points[:, 0]) y_max = np.max(points[:, 1]) w = x_max - x_min h = y_max - y_min wn = w // pixel_size + 1 hn = h // pixel_size + 1 x_bins = np.array([i * pixel_size for i in range(int(wn + 1))]) y_bins = np.array([i * pixel_size for i in range(int(hn + 1))]) df = pd.DataFrame(points, columns=['x', 'y', 'z', 'classification', 'indices']) df['x_bin'] = pd.cut(df['x'], bins=x_bins) df['y_bin'] = pd.cut(df['y'], bins=y_bins) result = df.groupby(['x_bin', 'y_bin']).apply(lambda x: x.loc[x['z'].idxmin()])[['x', 'y', 'z', 'classification', 'indices']] return result a = np.random.random([100,3]) b = np.random.random([100, 1])//0.5 c = np.arange(1000).reshape([100,1]) a = np.concatenate([a,b,c],axis=-1) d = localmin(a,0.2) 如何从a中删除d中的点

可以使用pandas的isin()方法来判断a中的点是否出现在d中，然后使用~运算符将不包含在d中的点筛选出来。具体实现代码如下：

import numpy as np 
import pandas as pd  

def localmin(points, pixel_size): 
    x_min = np.min(points[:, 0]) 
    y_min = np.min(points[:, 1]) 
    x_max = np.max(points[:, 0]) 
    y_max = np.max(points[:, 1]) 
    w = x_max - x_min 
    h = y_max - y_min 
    wn = w // pixel_size + 1 
    hn = h // pixel_size + 1 
    x_bins = np.array([i * pixel_size for i in range(int(wn + 1))]) 
    y_bins = np.array([i * pixel_size for i in range(int(hn + 1))]) 
    df = pd.DataFrame(points, columns=['x', 'y', 'z', 'classification', 'indices']) 
    df['x_bin'] = pd.cut(df['x'], bins=x_bins) 
    df['y_bin'] = pd.cut(df['y'], bins=y_bins) 
    result = df.groupby(['x_bin', 'y_bin']).apply(lambda x: x.loc[x['z'].idxmin()])[['x', 'y', 'z', 'classification', 'indices']] 
    return result 

a = np.random.random([100,3]) 
b = np.random.random([100, 1])//0.5 
c = np.arange(1000).reshape([100,1]) 
a = np.concatenate([a,b,c],axis=-1) 
d = localmin(a,0.2)

# 筛选出不包含在d中的点并重新赋值给a
a = a[~a[:, :2].isin(d.iloc[:, :2].values).any(axis=1)]

其中，isin()方法用于判断一个DataFrame或Series对象中的值是否在另一个DataFrame或Series对象中出现过。在这里我们用isin()方法来判断a中的前两列（即x和y坐标）是否出现在d中，然后使用any()方法将结果沿着列方向合并，得到一个长度为100的布尔型数组，表示a中每个点是否在d中出现过。最后使用~运算符对数组进行取反，得到不包含在d中的点的布尔型数组，然后使用该数组对a进行筛选即可。