Question

mainArray=np.linspace(1,50,50)
##group1 is first 10 elements of mainArray    
group1=np.array(1,2,3...10)
group2=np.array(11,12,13...20)
group3=np.array(21,22,23...30)
.
.
group5=np.array(41,42,43,44,45,46,47,48,49,50)
#i need to find standart deviation and mean value of these groups
#like np.mean(group1) and np.std(group1) for all groups
#then i have to calculate (group1-meanOfGroup1)/stdOfGroup1 for all groups
#and append it to one list or array.

我不知道如何通过循环解决这个问题，因为我的主要问题是我无法创建一个循环，将mainArray分组并应用np.mean和np.std。

Answer 1

根据您的上一条评论，您可以执行以下操作：

# create a list whith values [1,2, ..., 100]
a = list(range(1, 101))
# Sum values by range of 3 values using list comprehension
final = [sum(a[k:k+3]) for k in range(0, len(a), 3)]
print(final)

输出：

[6,
 15,
 24,
 33,
 42,
 51,
 60,
 ...
 267,
 276,
 285,
 294,
 100]

PS：最后一笔金额等于100，因为在1 to 100之间加上3个值的元素会将它们分组为：1->3, 4->6, ..., 94->96, 97->99, and the final element will be only one number which is 100

修改

使用list slicing进行上次编辑的实现，如下所示。

例如：

a = np.linspace(1,50,50) # unpacking groups group1, group2, group3, group4, group5 = [a[k:k+10] for k in range(0, len(a), 10)] # Applying: np.mean() to the groups print(np.mean(group1)) print(np.mean(group2)) print(np.mean(group3)) print(np.mean(group4)) print(np.mean(group5))

输出：

5.5 15.5 25.5 35.5 45.5

Answer 2

oc=np.linspace(1,100,100)
ocmean= [np.mean(oc[k:k+25]) for k in range(0,len(oc),25)]
ocstdev=[np.std(oc[k:k+25]) for k in range(0,len(oc),25)]  
## this 2 loops works, i got ocmean and ocstdev.
d=[(oc[k:k+25]-ocmean[i])/ocstdev[i] for k in range(0,len(oc),25) for i in range(len(ocmean))]
## This "d" doesnt gives what i want. It gives a list with 16*25 elements.

Answer 3

ocmean= [np.mean(oc[k:k+25]) for k in range(0,len(oc),25)]
ocstdev=[np.std(oc[k:k+25]) for k in range(0,len(oc),25)]     
d=[]

i=0
k=0
while i <len(ocmean):
    m=(oc[k:k+25]-ocmean[i])/ocstdev[i]
    i=i+1
    k=k+25
    d.append(m)

像这样解决

如何用循环切片数组并对它们应用操作？

3 个答案: