Where is the bottleneck in my Cython code?

Question

The profile tells me it took ~15s to run, but without telling me more.

Tue Aug 19 20:55:38 2014 Profile.prof
3 function calls in 15.623 seconds
Ordered by: internal time
ncalls tottime percall cumtime percall filename:lineno(function)
 1 15.623 15.623 15.623 15.623 {singleLoan.genLoan}
 1 0.000 0.000 15.623 15.623 <string>:1(<module>)
 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}

import numpy as np
cimport numpy as np
from libc.stdlib cimport malloc, free
from libc.stdlib cimport rand, srand, RAND_MAX
import cython
cimport cython
import StringIO
cdef extern from "math.h":
 int floor(double x)
 double pow(double x, double y)
 double exp(double x)
cdef double[:] zeros = np.zeros(360)
cdef double[:] stepCoupons = np.array([2.0,60,3.0,12.0,4.0,12.0,5.0])
cdef double[:,:] zeros2 = np.empty(shape=(999,2))
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1 10 0 0 fo 742 0.10162
1 10 0 0 fo 743 0.10846
1 10 0 0 fo 744 0.11531
1 10 0 0 fo 745 0.12216
1 10 0 0 fo 746 0.12902
1 10 0 0 fo 747 0.13588
1 10 0 0 fo 748 0.14274
1 10 0 0 fo 749 0.1496
1 10 0 0 fo 750 0.15646
1 10 0 0 xcumC 0 -0.64115
1 10 0 0 xcumC 1 -0.622
1 10 0 0 xcumC 2 -0.60285
1 10 0 0 xcumC 3 -0.5837
1 10 0 0 xcumC 4 -0.56455
1 10 0 0 xcumC 5 -0.54541
1 10 0 0 xcumC 6 -0.52628
1 10 0 0 xcumC 7 -0.50716
1 10 0 0 xcumC 8 -0.48805
1 10 0 0 xcumC 9 -0.46896
1 10 0 0 xcumC 10 -0.4499
1 10 0 0 xcumC 11 -0.43086
1 10 0 0 xcumC 12 -0.41186
1 10 0 0 xcumC 13 -0.39291
1 10 0 0 xcumC 14 -0.374
1 10 0 0 xcumC 15 -0.35515
1 10 0 0 xcumC 16 -0.33636
1 10 0 0 xcumC 17 -0.31764
1 10 0 0 xcumC 18 -0.29899
1 10 0 0 xcumC 19 -0.28044
1 10 0 0 xcumC 20 -0.26197
1 10 0 0 xcumC 21 -0.24361
1 10 0 0 xcumC 22 -0.22536
1 10 0 0 xcumC 23 -0.20722
1 10 0 0 xcumC 24 -0.18921
1 10 0 0 xcumC 25 -0.17133
1 10 0 0 xcumC 26 -0.1536
1 10 0 0 xcumC 27 -0.13602
1 10 0 0 xcumC 28 -0.11859
1 10 0 0 xcumC 29 -0.10134
1 10 0 0 xcumC 30 -0.08426
1 10 0 0 xcumC 31 -0.06736
1 10 0 0 xcumC 32 -0.05066
1 10 0 0 xcumC 33 -0.03416
1 10 0 0 xcumC 34 -0.01786
1 10 0 0 xcumC 35 -0.00179
''')
matParaC2P = np.genfromtxt(paraC2P, names= ['pt1','ct','mods','mbas','para','x','y'], dtype=['<f8','<f8','<f8','<f8','|S30','<f8','<f8'], delimiter='\t')
cdef packed struct paraArray:
 double pt1
 double ct
 double mods
 double mbas
 char[30] para
 double x
 double y
@cython.boundscheck(False)
@cython.wraparound(False)
@cython.nonecheck(False)
@cython.cdivision(True)
cpdef genc2p(paraArray [:] mat_p = matParaC2P, double pt_p= 1.0, double ct_p=10.0, double mbas_p =0.0, double mods_p=0.0, double count_c_p=24.0, double lv_p=60.0, double fo_p=740.0, double incentive_p = 200.0):
 cdef double[:,:] lvsubmat = getSubParaMat(input_s = mat_p, para_s='lv', ct_s = ct_p, pt1_s=pt_p, mbas_s=mbas_p,mods_s=0.0)
 cdef double lv0 = interp2d(lvsubmat[:,0], lvsubmat[:,1], lv_p)
 cdef double[:,:] fosubmat = getSubParaMat(input_s = mat_p, para_s='fo', ct_s = ct_p, pt1_s=pt_p, mbas_s=mbas_p,mods_s=0.0)
 cdef double fo0 = interp2d(fosubmat[:,0], fosubmat[:,1], fo_p)
 cdef double[:,:] cumCsubmat = getSubParaMat(input_s = mat_p, para_s='xcumC', ct_s = ct_p, pt1_s=pt_p, mbas_s=mbas_p,mods_s=0.0)
 cdef double cumC0 = interp2d(cumCsubmat[:,0], cumCsubmat[:,1], count_c_p)
 cdef double[:,:] incsubmat = getSubParaMat(input_s= mat_p, para_s='dollarSaving', ct_s = ct_p, pt1_s=pt_p, mbas_s=mbas_p,mods_s=mods_p)
 cdef double inc0 = interp2d(incsubmat[:,0], incsubmat[:,1], incentive_p)
 cdef double[:,:] intercept = getSubParaMat(input_s= mat_p, para_s='(Intercept)', ct_s = ct_p, pt1_s=pt_p, mbas_s=mbas_p,mods_s=0.0)
 cdef double intercept0 = intercept[0,1]
 return genLogit(lv0+fo0+cumC0+inc0+intercept0)
 #return intercept0
cpdef double[:,:] getSubParaMat(paraArray [:] input_s = matParaC2P, double pt1_s =1.0, double ct_s=10.0, double mods_s=0.0, double mbas_s=0.0, char[30] para_s = 'lv'):
 cdef int start = 0, end =0, k=0
 cdef double[:,:] output = zeros2
 for i from 0<=i<input_s.size:
 if input_s[i].pt1 == pt1_s and input_s[i].ct == ct_s and input_s[i].mods==mods_s and input_s[i].mbas==mbas_s and input_s[i].para[:]==para_s[:] and start==0:
 start = i
 end = i
 while end<input_s.size and input_s[end].pt1 == pt1_s and input_s[end].ct == ct_s and input_s[end].mods==mods_s and input_s[end].mbas==mbas_s and input_s[end].para[:]==para_s[:]:
 (output[k,0], output[k, 1]) = (input_s[end].x, input_s[end].y)
 end += 1
 k+=1
 break
 return output[:k, :]
cpdef double interp2d(double[:] x, double[:] y, double new_x, int ex = 0):
 cdef int nx = x.shape[0]-1
 cdef int ny = y.shape[0]-1
 cdef double new_y
 cdef int steps=0
 if ex==0 and new_x<x[0]:
 new_x = x[0]
 elif ex==0 and new_x>x[nx]:
 new_x = x[nx]
 if new_x<=x[0]:
 new_y = (new_x - x[0])*(y[0]-y[1])/(x[0] - x[1]) + y[0]
 elif new_x>=x[nx]:
 new_y = (new_x - x[nx])*(y[ny] - y[ny-1])/(x[nx] - x[nx-1]) + y[ny]
 else:
 while new_x>x[steps]:
 steps +=1
 new_y = (new_x - x[steps-1])*(y[steps] - y[steps-1])/(x[steps] - x[steps-1]) + y[steps-1]
 return new_y
cpdef double genLogit(double total):
 return 1.0/(1.0+exp(-1.0*(total)))
cpdef genLoan():
 for j from 0<=j<100:
 for i from 1<=i<240:
 prob_p = genc2p(mat_p = matParaC2P)

• Schism: I don't use cython, but that's a terrible profiler if I've ever seen one. Is there a different one you can use?
• Schism: I'll note that genc2p is called 24 000 times, so interp2d is being called 96 000 times and getSubParaMat is called 120 000 times. You're also iterating through input_s in each call of getSubParaMat, and you have nested while loops in each of those. I'm sorry I can't help any more though.
• ali_m: It would be really helpful if you could prune this example down a bit. Also, the example string input you give is not copy-pasteable, since SE converts tabs to spaces.

Answer 1

There's a lot of code here, so I'm just going to review interp2d.

1. There's no docstring. What does this function do? How am I supposed to call it? Are there any constraints on the parameters (for example, does the x array need to be sorted)?.

2. The function seems to be misnamed: it interpolates the function that takes x to y, but this is a function of one argument, so surely interp1d or just interp would be better names. (Compare numpy.interp, which does something very similar to your function, but is documentated as "one-dimensional linear interpolation".)

3. The parameter ex is opaquely named. What does it mean? Reading the code, it seems that it controls whether or not to extrapolate for values of x outside the given range. So it should be named extrapolate and it should be a Boolean (True or False) not a number.

4. Python allows indexing from the end of an array, so you can write x[-1] for the last element of an array, instead of the clusmy x[x.shape[0] - 1].

5. You find the interval in which to do the interpolation by a linear search over the array x, which takes time proportional to the size of x and so will be slow when x is large. Since x needs to be sorted in order for this algorithm to make sense, you should use binary search (for example, numpy.searchsorted) so that the time taken is logarithmic in the size of x.

6. You have failed to vectorize this function. The whole point of NumPy is that it provides fast operations on arrays of fixed-size numbers. If you find yourself writing a function that operates on just one value at a time (here, the single value new_x) then you probably aren't getting much, if any, benefit from NumPy.

7. Putting all this together, I'd write something like this:

   def interp1d(x, y, a, extrapolate=False):
    """Interpolate a 1-D function.
    x is a 1-dimensional array sorted into ascending order.
    y is an array whose first axis has the same length as x.
    a is an array of interpolants.
    If extrapolate is False, clamp all interpolants to the range of x.
    Let f be the piecewise-linear function such that f(x) = y.
    Then return f(a).
    >>> x = np.arange(0, 10, 0.5)
    >>> y = np.sin(x)
    >>> interp1d(x, y, np.pi)
    0.0018202391415163
    """
    if not extrapolate:
    a = np.clip(a, x[0], x[-1])
    i = np.clip(np.searchsorted(x, a), 1, len(x) - 1)
    j = i - 1
    xj, yj = x[j], y[j]
    return yj + (a - xj) * (y[i] - yj) / (x[i] - xj)
   

   (The rest of the program may need to be reorganized to pass arrays of values instead of one value at a time.)

8. Finally, what was wrong with scipy.interpolate.interp1d? It doesn't provide quite the same handling of points outside the range, but you could call numpy.clip yourself before calling it.