elegant way of convert a numpy array containing datetime.timedelta into seconds in python 2.7
Question: I have a numpy array called `dt`. Each element is of type
`datetime.timedelta`. For example:

    
    
    >>>dt[0]
    datetime.timedelta(0, 1, 36000)
    

how can I convert `dt` into the array `dt_sec` which contains only seconds
without looping? my current solution (which works, but I don't like it) is:

    
    
    dt_sec = zeros((len(dt),1))
    for i in range(0,len(dt),1):
        dt_sec[i] = dt[i].total_seconds()
    

I tried to use `dt.total_seconds()` but of course it didn't work. any idea on
how to avoid this loop?

Thanks


Answer: 
    import numpy as np
    
    helper = np.vectorize(lambda x: x.total_seconds())
    dt_sec = helper(dt)
    

