CODEWITHSUNDEEP
pythonpandasdatetimedataframe
Pau
Pau

Reputation: 362

How can I convert a single integer representing days into pandas datetime

I have a dataframe with a column "time" of float numbers, representing days from 0 to 8, and one more column with other data. The time step is not continuous.

time_clean = np.arange(0, 8, 0.1)
noise = [random.random()/10 for n in range(len(time_clean))]
time = time_clean + noise

data = [random.random()*100 for n in range(len(time_clean))]

df = pd.DataFrame({"time": time, "data":data})
df.head()

      data      time
0  89.965240  0.041341
1  95.964621  0.109215
2  70.552763  0.232596
3  74.457244  0.330750
4  13.228426  0.471623

I want to resample and interpolate the data to every 15 minutes, (15/(60*24) days).

I think the most efficient way to do this would be using the resample method of pandas dataframes, but in order to do that I need to convert the time column into a timestamp, and make it the index.

What is the most efficient way of doing this? Is it possible to transform an int to datetime?

Upvotes: 3

Views: 133

Answers (1)

jezrael
jezrael

Reputation: 862711

I think you need first convert column time to_timedelta and then sort_values with resample:

Also I think the best is add one new row with 0 for always starts resample from 0 (if 0 is not in time column it starts from minimal time value)

df.loc[-1] = 0
df.time = pd.to_timedelta(df.time, unit='d')
df = df.sort_values('time').set_index('time').resample('15T').ffill()
print (df.head(20))
               data
time               
00:00:00   0.000000
00:15:00   0.000000
00:30:00   0.000000
00:45:00   0.000000
01:00:00   0.000000
01:15:00   0.000000
01:30:00  50.869889
01:45:00  50.869889
02:00:00  50.869889
02:15:00  50.869889
02:30:00  50.869889
02:45:00  50.869889
03:00:00  50.869889
03:15:00   8.846017
03:30:00   8.846017
03:45:00   8.846017
04:00:00   8.846017
04:15:00   8.846017
04:30:00   8.846017
04:45:00   8.846017

Upvotes: 3

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