Question

想知道是否有人可以帮助我解决一些我遇到的小问题。

list1 = [1,2,4]
dz = 0.5
k = max(list1)/dz
print('k =',k)
list1_diff = np.insert(np.diff(list1),0,1)
for item in list1_diff:
    #Number of times to repeat number
    repeats = item/dz
    vec = [np.random.normal(0,1)]*int(repeats)
    print(vec)

这产生

[-0.7014481088047718, -0.7014481088047718]
[3.1264015795601274, 3.1264015795601274]
[0.44017976879654314, 0.44017976879654314, 0.44017976879654314, 0.44017976879654314]

这在做些正确的事情，但是我希望循环的输出是将这三个列表作为一个列表。所以应该是

[-0.7014481088047718, -0.7014481088047718, 3.1264015795601274, 3.1264015795601274, 0.44017976879654314, 0.44017976879654314, 0.44017976879654314, 0.44017976879654314]

Answer 1

你不傻。如果我答对了，那么您希望根据您的公式创建一个包含重复随机数的列表。您只需要修改一下代码，就像这样：

list1 = [1,2,4]
dz = 0.5
k = max(list1)/dz
print('k =',k)
list1_diff = np.insert(np.diff(list1),0,1)
# create your empty result list, where you add the random numbers in the loop
vec= list()
for item in list1_diff:
    # Number of times to repeat number
    repeats = item/dz
    # add the new random number with it's repeats to the vec list
    # using extend
    vec.extend([np.random.normal(0,1)]*int(repeats))
# print the result outside the loop, after it has been constructed
print(vec)

Answer 2

尝试这个：

import numpy as np

list1 = [1,2,4]
dz = 0.5
k = max(list1)/dz
print('k =',k)
list1_diff = np.insert(np.diff(list1),0,1)
final_list=[]
for item in list1_diff:
    #Number of times to repeat number
    repeats = item/dz
    vec = [np.random.normal(0,1)]*int(repeats)
    final_list += vec
print(final_list)

Answer 3

这应该可以完成工作：

import numpy as np

list1 = [1,2,4]
dz = 0.5
k = max(list1)/dz
print(f'k = {k}')

lst = [[np.random.normal(0,1)]*int(item/dz) for item in np.insert(np.diff(list1),0,1)]
flatten = lambda l: [item for sublist in l for item in sublist]

print(flatten(lst))

输出：

k = 8.0
[-1.1762058341816053, -1.1762058341816053, -1.0599923573472354, -1.0599923573472354, 0.6374252888036466, 0.6374252888036466, 0.6374252888036466, 0.6374252888036466]

这里发生了什么

列表理解可创建一个完整的输出列表（主要是您所做的事情）
拉平函数应用于列表

请注意，您也可以使用extend()，我打算将其添加到其中，但另一个答案击败了我

Answer 4

Python列表对此具有内置函数：

    String date = "2020-09-28T11:47:37.217";
    int from = date.indexOf("T")+1;
    int to = from+5;
    
    String timeOfInterest = date.substring(from, to);
    System.out.println( timeOfInterest );

例如，

list_name.extend([])

[1、2、3、4、5、6]

因此python extend 函数将列表扩展为另一个列表。它不会附加它！

在for循环中创建列表

4 个答案: