如何检查病情是否持续超过15分钟? [英] How to check if a condition lasted for more than 15 mins?

问题描述

下面是数据集的示例

<身体>

日期	值
2020-01-01 01:35	50
2020-01-01 01:41	49
2020-01-01 01:46	50

我希望检查连续15分钟的值"是否等于50.如果是，我想提取它发生的日期.让我举一个例子，我说连续15分钟.假设我要在5分钟(而不是15分钟)的连续时间内检查该值是否等于50.满足该条件的数据如下

I wish to check if the 'Value' was equal to 50 for continuous period of 15 mins. If yes, I want to extract the date for which it occurred. Let me give an example what I mean by continuous period of 15 mins. Assume that I want to check if the value is equal to 50 for a continuous period of 5 mins (instead of 15 mins). The data that would satisfy this condition would be as follows

<身体>

日期	值
2020-01-01 01:35	50
2020-01-01 01:36	50
2020-01-01 01:37	50
2020-01-01 01:38	50
2020-01-01 01:39	50

然后我想将日期 2020-01-01 提取到列表中，因为上述数据连续5分钟(或更长)等于50.

Then I want to extract the date2020-01-01 onto a list because the above data was equal to 50 for a continuous period of 5 mins (or more).

推荐答案

我将代码发布5分钟，以便输出与您所需的输出匹配.将 300 更改为 900 15分钟.步骤:

I am posting code for 5 mins so that output matches your desired output. Change 300 to 900 for 15 mins. Steps:

1. 将 df ['Date'] 转换为 datetime ，以便我们可以减去两个日期知道他们之间的时差.

1. Convert the df['Date'] to datetime so that we can subtract two dates to know the time difference between them.

按日期对 df 进行分组，并为每个分组对象调用 f .

Group the df by date and Call f for each group object.

在 f 中: max-continuous_range 给出了长度为50的最长段的长度.如果长度为5分钟或以上，则 f 返回True.如果 f 返回 True ，则在列表中追加日期.

In f: max-continuous_range gives the length of longest segment where value is 50. f return True if length is 5 mins or more. Append date in list if f returns True.

使用:

def f(g):
   mask = (g['Value'] == 50)
   max_continuous_range = (np.max(np.cumsum(g['Date'].where(mask).diff()))
                         + timedelta(minutes = 1))
   return  max_continuous_range.seconds >= 300

df['Date'] = pd.to_datetime(df['Date'])
groups = df.groupby(df['Date'].dt.date, as_index = False)
final_list = [str(idx) for idx, g in groups if f(g)]

输入:

    Date  Value
0  2020-01-01 01:35     40
1  2020-01-01 01:36     50
2  2020-01-01 01:37     50
3  2020-01-01 01:38     50
4  2020-01-01 01:39     50
5  2020-01-01 01:40     50
6  2020-01-01 01:41     40
7  2020-01-01 01:42     40

输出:

>>> final_list
['2020-01-01']

在f(g)内:

掩码:真，值是50.

    0    False
    1     True
    2     True
    3     True
    4     True
    5     True
    6    False
    7    False

df ['Date'].where(mask)将NaT放在mask不是True的地方.

df['Date'].where(mask) Puts NaT where mask is not True.

0                   NaT
1   2020-01-01 01:36:00
2   2020-01-01 01:37:00
3   2020-01-01 01:38:00
4   2020-01-01 01:39:00
5   2020-01-01 01:40:00
6                   NaT
7                   NaT

.diff 给出两个连续元素之间的区别.如果任何值为NaT，它将给出NaT. df ['Date'].where(mask).diff():

.diff gives difference between two consecuting elements. It will give NaT if any value is NaT. Result after df['Date'].where(mask).diff():

0               NaT
1               NaT
2   0 days 00:01:00
3   0 days 00:01:00
4   0 days 00:01:00
5   0 days 00:01:00
6               NaT
7               NaT

现在，连续时间之间的累计差值总和将为我们提供经过的总时间.在 np.cumsum(...)之后:

Now cumulative sum of difference between consecutive times will give us the total time elapsed. After np.cumsum(...):

0               NaT
1               NaT
2   0 days 00:01:00
3   0 days 00:02:00
4   0 days 00:03:00
5   0 days 00:04:00
6               NaT
7               NaT

np.max 给了我们最长的长度.添加 1 分钟以处理边界条件

np.max gives us the longest length. 1 minute is added to take care of boundary condition

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