Aggregating time series data

Question

I am no data scientist. I do know python and I currently have to manage time series data that is coming in at a regular interval. Much of this data is all zero's or values that are the same for a long time, and to save memory I'd like to filter them out. Is there some standard method for this (which I am obviously unaware of) or should I implement my own algorithm?

What I want to achieve is the following:

interval  value   result
(summed) 
1         0       0
2         0       # removed
3         0       0
4         1       1
5         2       2
6         2       # removed
7         2       # removed
8         2       2
9         0       0
10        0       0

Any help appreciated !

Answer 1

You could use pandas query on dataframes to achieve this:

import pandas as pd

matrix = [[1,0, 0],
[2, 0, 0],
[3, 0, 0],
[4, 1, 1],
[5, 2, 2],
[6, 2, 0],
[7, 2, 0],
[8, 2, 2],
[9, 0, 0],
[10,0, 0]]


df = pd.DataFrame(matrix, columns=list('abc'))
print(df.query("c != 0"))

Answer 2

There is no quick function call to do what you need. The following is one way

import pandas as pd

df = pd.DataFrame({'interval':[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
             'value':[0, 0, 0, 1, 2, 2, 2, 2, 0, 0]}) # example dataframe


df['group'] = df['value'].ne(df['value'].shift()).cumsum() # column that increments every time the value changes

df['key'] = 1 # create column of ones
df['key'] =  df.groupby('group')['key'].transform('cumsum') # get the cumulative sum 

df['key'] = df.groupby('group')['key'].transform(lambda x: x.isin( [x.min(), x.max()])) # check which key is minimum and which is maximum by group

df = df[df['key']==True].drop(columns=['group', 'key']) # keep only relevant cases

df

Answer 3

Here is the code:

l = [0, 0, 0, 1, 2, 2, 2, 2, 0, 0]

for (i, ll) in enumerate(l):
    if i != 0 and ll == l[i-1] and i<len(l)-1 and ll == l[i+1]: 
        continue
    print(i+1, ll)

It produces what you want. You haven't specified format of your input data, so I assumed they're in a list. The conditions ll == l[i-1] and ll == l[i+1] are key to skipping the repeated values.

Answer 4

Thanks all. Looking at the answers I guess I can conclude I'll need to roll my own. I'll be using your input as inspiration. Thanks again !