Use of datetime timedelta with numpy 3d array

Question

I have a 3D array with the count of number of days past a benchmark date (e.g., 01.01.2000). I am interested in the actual day-of-year (DOY: 1-365/366)rather than the total number of days past a given date.

For a single value, the below syntax works. For e.g.,

import numpy as np
import datetime

data = 1595
date = datetime.datetime(2000,1,1,0,0) + datetime.timedelta(data -1)
date.timetuple().tm_yday
134

However, I am having issues with using a 3D array.

import numpy as np
import datetime

data = np.random.randint(5, size = (2,2,2))
data = data + 1595
data
array([[[1596, 1595],
    [1599, 1599]],

   [[1596, 1599],
    [1595, 1595]]])

#Function
def Int_to_DOY(int_array):
    date_ = datetime.datetime(2000,1,1,0,0) + datetime.timedelta(int_array - 1)
    return date_.timetuple().tm_yday

doy_data = data * 0 #Empty array

for i in range(2):
    doy_data[:, :, i] = Int_to_DOY(data[:, :, i])

Here is the error message and I am not able to figure this out.

TypeError: unsupported type for timedelta days component: numpy.ndarray

Thanks for your help.

array([[[135, 134], [138, 138]], [[135, 138], [134, 134]]]) — pyPN
– pyPN, Commented Sep 10, 2019 at 14:07

user2874583 · Accepted Answer · 2019-09-10 14:08:42Z

1

import numpy as np
import datetime

data = np.random.randint(5, size = (2,2,2))
data = data + 1595

#Function
def Int_to_DOY(int_array):
    date_ = datetime.datetime(2000,1,1,0,0) + datetime.timedelta(int(int_array) -1)
    return date_.timetuple().tm_yday

doy_data = data.flatten()

for i in range(len(doy_data)):
    doy_data[i] = Int_to_DOY(doy_data[i])

doy_data = doy_data.reshape((2,2,2))

answered Sep 10, 2019 at 14:08

user2874583

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1 Comment

pyPN Over a year ago

Thank you for the solution and I believe, I was pushing a 2D array wherein it should have been a vector.

Quang Hoang · Accepted Answer · 2019-09-10 14:08:40Z

1

Since you tagged pandas:

data = np.array([[[1596, 1595],
    [1599, 1599]],

   [[1596, 1599],
    [1595, 1595]]])

s = pd.to_datetime('2000-01-01') + pd.to_timedelta(data.ravel(), unit='D') 
s.dayofyear.values.reshape(data.shape) - 1

Output:

array([[[135, 134],
        [138, 138]],

       [[135, 138],
        [134, 134]]], dtype=int64)

answered Sep 10, 2019 at 14:08

Quang Hoang

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2 Comments

pyPN Over a year ago

Thank you! I like the pandas approach and something new I learnt today.

pyPN Over a year ago

Note: On a larger array, pandas is way faster!

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