Question

我有一个包含2d数组的字典。我试图通过这种方式计算均值，但它不起作用，因为数组也包含nan值。那么有没有更简单的方法来计算平均值？

All = np.zeros(385000).reshape(550,700)

for i  in dic.keys():
    a = dic[i]['data']
    avg = (All+a)/len(dic.keys())

Answer 1

您似乎正在尝试在输入a和b中逐个考虑元素，忽略NaNs。因此，一种方法是使用np.dstack堆叠这两个数组，这将沿着第三个轴堆叠a和b，然后沿着同一轴简单地使用np.nanmean。因此，我们将有一个像这样的简单实现 -

np.nanmean(np.dstack((a,b)),axis=2)

示例运行 -

In [28]: a
Out[28]: 
array([[  2.,  nan],
       [  5.,   4.]])

In [29]: b
Out[29]: 
array([[ nan,   3.],
       [  7.,   2.]])

In [30]: np.nanmean(np.dstack((a,b)),axis=2)
Out[30]: 
array([[ 2.,  3.],
       [ 6.,  3.]])

对于从问题的发布代码中显示的字典中获取那些2D数组的情况，您可以使用循环理解将这些数组收集为3D数组使用np.dstack并最后沿最后一个轴使用np.nanmean，就像这样 -

np.nanmean(np.dstack([d['data'] for d in dic]),axis=2)

Answer 2

上面的答案是绝对好的，但是version: '3.1' services: cassandra: container_name: "cassandra" image: cassandra ports: - 9042:9042 volumes: - /home/cassandra:/var/lib/cassandra postgresql: container_name: "postgresql" image: postgres:11.1-alpine restart: always environment: POSTGRES_DB: mywebapp POSTGRES_USER: postgres POSTGRES_PASSWORD: postgres volumes: #- ./startup.sql:/docker-entrypoint-initdb.d/startup.sql - postgresdata:/var/lib/postgresql/data ports: - 5432:5432 mywebapp: container_name: "mywebapp" image: openjdk:10-jre-slim hostname: mywebapp volumes: - ./lib:/home/lib - ./mywebapp-1.0.1-SNAPSHOT-exec.jar:/home/mywebapp-1.0.1-SNAPSHOT-exec.jar entrypoint: - java - -jar - -Djava.library.path=/home/lib - /home/mywebapp-1.0.1-SNAPSHOT-exec.jar environment: - LD_LIBRARY_PATH=/home/lib - spring.datasource.url=jdbc:postgresql://postgresql:5432/mywebapp - spring.cassandra.contactpoints=cassandra - spring.cassandra.port=9042 - spring.cassandra.keyspace=mywebapp #- spring.datasource.username=postgres #- spring.datasource.password=postgres #- spring.jpa.hibernate.ddlAuto=update+ ports: - 8443:8443 - 8080:8080 depends_on: - cassandra似乎不太灵活或直观。我们还可以使用np.dstack((a,b))来提供更直观的显示。请参见下面的示例。

代码

np.stack()

输出

a=np.array([[2,np.nan],[5,4]])
b=np.array([[np.nan,3],[7,2]])
c=np.stack((a,b),axis=0)
print(a)
print('='*50)
print(b)
print('='*50)
print(c)
print('='*50)
print(np.nanmean(c,axis=0))

[[ 2. nan] [ 5. 4.]] ================================================== [[nan 3.] [ 7. 2.]] ================================================== [[[ 2. nan] [ 5. 4.]] [[nan 3.] [ 7. 2.]]] ================================================== [[2. 3.] [6. 3.]]和np.dstack()之间的差异可以通过我编写的以下示例找到。

代码

np.stack()

输出

dr1=np.array([[1,2,3],[4,5,6],[7,8,9]])
print(dr1)
dr2=np.array([[9,8,7],[6,5,4],[3,2,1]])
print(dr2)
print('='*50)
dr3=np.dstack((dr1,dr2))
print(dr3.shape)
print(dr3)
print(np.sum(dr3,axis=2)) # This will be (row,col,time) but display (col,time) => (row,col) in each row, the 1 in dr2 will be in [3,3,2] => 3 [3,2]
print('='*50)
dr4=np.stack((dr1,dr2),axis=0) # This will be (time,row,col) and display (row,col) => (row,col) in each time, the 1 in dr2 will be in [2,3,3] => 2 [3,3]
print(dr4.shape)
print(dr4)
print(np.sum(dr4,axis=0))

如何计算2d数组的np.nanmean

2 个答案:

代码

输出

代码

输出