I try to rewrite a complex function form Python to Cython to speed it up massively and I encounter the following problem: while compiling my function hh_vers_vector.pyx using
setup(
ext_modules=cythonize("hh_vers_vector.pyx"),
)
it throws the following error
cdef int numSamples = len(Iext);
# initial values
cdef float v[numSamples]
^
------------------------------------------------------------
hh_vers_vector.pyx:47:27: Not allowed in a constant expression
If, however, I give "numSamples" as a number into the function, there is no problem. I don't understand that, because I thought that len(Iext) will return a number as 10.000 as well. So why does Cython have a problem with this expression?
cdef float v[numSamples] # ERROR
cdef float v[10000] # NO ERROR
My function looks like this so far:
from math import exp
import time
def hhModel(*params, Iext, float dt, int Vref):
## Unwrap params argument: these variables are going to be optimized
cdef float ENa = params[0]
cdef float EK = params[1]
cdef float EL = params[2]
cdef float GNa = params[3]
cdef float GK = params[4]
cdef float GL = params[5]
## Input paramters
# I : a list containing external current steps, your stimulus vector [nA/1000]
# dt : a crazy time parameter [ms]
# Vref : reference potential [mV]
# T : Total simulation time in [ms]
def alphaM(float v, float vr): return 0.1 * (v-vr-25) / ( 1 - exp(-(v-vr-25)/10) )
def betaM(float v, float vr): return 4 * exp(-(v-vr)/18)
def alphaH(float v, float vr): return 0.07 * exp(-(v-vr)/20)
def betaH(float v, float vr): return 1 / ( 1 + exp( -(v-vr-30)/10 ) )
def alphaN(float v, float vr): return 0.01 * (v-vr-10) / ( 1 - exp(-(v-vr-10)/10) )
def betaN(float v, float vr): return 0.125 * exp(-(v-vr)/80)
## steady-state values and time constants of m,h,n
def m_infty(float v, float vr): return alphaM(v,vr) / ( alphaM(v,vr) + betaM(v,vr) )
def h_infty(float v, float vr): return alphaH(v,vr) / ( alphaH(v,vr) + betaH(v,vr) )
def n_infty(float v, float vr): return alphaN(v,vr) / ( alphaN(v,vr) + betaN(v,vr) )
## parameters
cdef float Cm, gK, gL, INa, IK, IL, dv_dt, dm_dt, dh_dt, dn_dt, aM, bM, aH, bH, aN, bN
cdef float Smemb = 4000 # [um^2] surface area of the membrane
cdef float Cmemb = 1 # [uF/cm^2] membrane capacitance density
Cm = Cmemb * Smemb * 1e-8 # [uF] membrane capacitance
gNa = GNa * Smemb * 1e-8 # Na conductance [mS]
gK = GK * Smemb * 1e-8 # K conductance [mS]
gL = GL * Smemb * 1e-8 # leak conductance [mS]
######### HERE IS THE PROBLEM ##############
cdef int numSamples = len(Iext);
######### HERE IS THE PROBLEM ##############
# initial values
cdef float v[numSamples]
cdef float m[numSamples]
cdef float h[numSamples]
cdef float n[numSamples]
v[0] = Vref # initial membrane potential
m[0] = m_infty(v[0], Vref) # initial m
h[0] = h_infty(v[0], Vref) # initial h
n[0] = n_infty(v[0], Vref) # initial n
## calculate membrane response step-by-step
for j in range(0, numSamples-1):
# ionic currents: g[mS] * V[mV] = I[uA]
INa = gNa * m[j]*m[j]*m[j] * h[j] * (ENa-v[j])
IK = gK * n[j]*n[j]*n[j]*n[j] * (EK-v[j])
IL = gL * (EL-v[j])
# derivatives
# I[uA] / C[uF] * dt[ms] = dv[mV]
dv_dt = ( INa + IK + IL + Iext[j]*1e-3) / Cm;
aM = 0.1 * (v[j]-Vref-25) / ( 1 - exp(-(v[j]-Vref-25)/10))
bM = 4 * exp(-(v[j]-Vref)/18)
aH = 0.07 * exp(-(v[j]-Vref)/20)
bH = 1 / ( 1 + exp( -(v[j]-Vref-30)/10 ) )
aN = 0.01 * (v[j]-Vref-10) / ( 1 - exp(-(v[j]-Vref-10)/10) )
bN = 0.125 * exp(-(v[j]-Vref)/80)
dm_dt = (1-m[j])* aM - m[j]*bM
dh_dt = (1-h[j])* aH - h[j]*bH
dn_dt = (1-n[j])* aN - n[j]*bN
# calculate next step
v[j+1] = (v[j] + dv_dt * dt)
m[j+1] = (m[j] + dm_dt * dt)
h[j+1] = (h[j] + dh_dt * dt)
n[j+1] = (n[j] + dn_dt * dt)
return v
