4
\$\begingroup\$

I have rewritten my python code in cython in order to speed up it quite a bit. However after checking the code in jupyter it seems that part of it is still compiled as python code, therefore its not being sped up enough.

My function is pretty basic, it gets start date, end date and creates array of additional dates based on that and some other conditions.

The thing is I’m a bit clueless how could I change the highlighted yellow part of code of the date_range_np function as seen on the attached imaged. Because I’m already using numpy arrays so I thought it would be compiled as cython code. There is also one loop that I think is slowing down the function but so far I wasn’t able to come up with any replacement that sginificatly sped up the function. Actually speed is very important as those functions are executed thousands of times. Any ideas how could I refactor it to speed it up?

Here is my code:

%%cython -a
import numpy as np
cimport numpy as np
from datetime import datetime

cpdef np.int64_t get_days(np.int64_t year, np.int64_t month):
    cdef np.ndarray months=np.array([31,28,31,30,31,30,31,31,30,31,30,31])
    if month==2:
        if (year%4==0 and year%100!=0) or (year%400==0):
            return 29
    return months[month-1]

cpdef np.ndarray[np.int64_t] date_range_np(np.int64_t start, np.int64_t end, char* freq):
    cdef np.ndarray res
    cdef np.int64_t m_start
    cdef np.int64_t m_end
    if freq.decode("utf-8")[len(freq)-1] == "M":
        m_start = np.int64(start).astype('M8[D]').astype('M8[M]').view("int64")
        m_end = np.int64(end).astype('M8[D]').astype('M8[M]').view("int64")
        res = np.arange(m_start, m_end-2, np.int64(freq[:(len(freq)-1)])).view("M8[M]").astype("M8[D]").view("int64")
        return np.array([np.min([x+datetime.fromtimestamp(start*24*60*60).day-1, x+get_days(datetime.fromtimestamp(start*24*60*60).year,datetime.fromtimestamp(start*24*60*60).month)]) for x in res])
    elif freq.decode("utf-8")[len(freq)-1] == "D":
        return np.arange(start, end-2, np.int64(freq[:(len(freq)-1)]))

cpdef np.ndarray[np.int64_t] loanDates(np.int64_t startDate,np.int64_t endDate,np.int64_t freq):
    # if frequency indicates repayment every n months
    if int(12 / freq) == 12 / freq:
        # Generate date range and add offset depending on starting day
        #print(date_range_np(start=startDate, end=endDate, freq=(str(-int(12 / freq)) + "M").encode("utf8")))
        ts = date_range_np(start=startDate, end=endDate, freq=(str(-int(12 / freq)) + "M").encode("utf8"))
    else:
        ts = date_range_np(start=startDate, end=endDate, freq=(str(-int(365 / freq)) + "D").enocode("utf8"))
    #print(ts)

    if ts.shape[0] == 0:
        return np.array([])
    elif ts.shape[0] >= 1 and ts[0] > startDate:
        ts = np.delete(arr=ts, obj=0)
    if ts[ts.shape[0]-1] < endDate:
        ts = np.delete(arr=ts, obj=-1)
    if ts[0] != startDate:
        ts = np.insert(ts,0,startDate)


    # If no dates generated (start date>end date)
    ts = ts

    # If last date generated is not end date add it
    return ts.astype('int64')

And the function highlighted:

bottleneck function

If you want to test the functions you can try:

%timeit date_range_np(20809,17986, b"-1M")

I get about 1ms for that function.

\$\endgroup\$
  • 1
    \$\begingroup\$ How does your function compare to pandas.date_range? \$\endgroup\$ – Graipher Aug 19 at 18:55
  • 1
    \$\begingroup\$ I used it before and it was slower \$\endgroup\$ – Alex T Aug 20 at 5:26
4
\$\begingroup\$

I'm sorry, but this code is really hard to read. I must admit I don't know Cython too well, so I won't be able to comment too much on that part. But anyways, here are a few comments, in random order.

  • While Cython does not fully support docstrings (they do not show up in the interactive help), this should not prevent you from adding some to explain what the different functions do and what arguments they take.

  • You seem to be doing np.int64(start).astype('M8[D]').astype('M8[M]').view("int64") quite a lot. As far as I can tell, this extracts the month from a date, which was given as an integer(?). There is quite possibly a better way to do that (using the functions in datetime), but they might be slower. Nevertheless, you should put this into its own function.

  • You do freq.decode("utf-8")[len(freq)-1] twice. Do it once and save it to a variable. Also, freq[len(freq)-1] should be the same as freq[-1] and freq[:len(freq)-1] the same as freq[:-1]. This is especially costly as len(freq) is \$\mathcal{O}(n)\$ for char *, in Cython.

  • You create datetime.fromtimestamp(start*24*60*60) three times, once each to get the day, month and year. Save it to a variable and reuse it.

  • The last two comments in loanDates seem not to be true anymore:

    # If no dates generated (start date>end date)
    ts = ts
    
    # If last date generated is not end date add it
    return ts.astype('int64')
    
  • The documentation seems to recommend against using C strings, unless you really need them. If I read the documentation correctly you could just make the type of freq str and get rid of all your encode("utf-8") and decode("utf-8") code.
  • The definition of the months array is done every time the function get_days is called. In normal Python I would recommend making it a global constant, here you would have to try and see if it makes the runtime worse.

  • Python has an official style-guide, PEP8. Since Cython is only an extension, it presumably also applies here. It recommends surrounding operators with whitespace (freq[len(freq) - 1]), using lower_case for all function and variable names and limiting your linelength (to 80 characters by default, but 120 is also an acceptable choice).

In the end, taking 1ms to create a date range is already quite fast. As you said this is already faster than pandas.daterange (which does a lot of parsing of the input first, which you avoid by passing in numbers directly). You might be able to push it down to microseconds, but you should ask yourself if and why you need this many dateranges per second.

\$\endgroup\$
  • 1
    \$\begingroup\$ Thanks for the reply. Are the points 2 and 4 suggested to save time or more to make the code more readable? \$\endgroup\$ – Alex T Aug 20 at 17:35
  • \$\begingroup\$ @AlexT Both! Although there is a small function overhead, it will probably be less than doing the computation twice (especially since it involves multiple attribute lookups). But when in doubt, measure it! \$\endgroup\$ – Graipher Aug 20 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.