How to increment global variable from function in python

Question

I'm stuck by a simple increment function like

from numpy import *
from pylab import *

## setup parameters and state variables
T       = 1000                # total time to simulate (msec)
dt      = 1                   # simulation time step (msec)
time    = arange(0, T+dt, dt) # time array
Vr      = -70                 #reset
El      = -70                

## LIF properties
Vm      = zeros(len(time))      # potential (V) trace over time 
Rm      = 10                    # resistance (mOhm)
tau_m   = 10                    # time constant (msec)
Vth     = -40                   # spike threshold (V)

## Input stimulus
I       = 3.1                 # input current (nA)
Vm[0] = -70

Fr = 0

## iterate over each time step
def func(Ie, Vm, Fr):
    for i, t in enumerate(time):
        if i == 0:
            Vm[i] = -70
        else: 
            Vm[i] = Vm[i-1] + (El- Vm[i-1] + Ie*Rm) / tau_m * dt
            if Vm[i] >= Vth:
                Fr += 1
                Vm[i] = El
     return

Ie = 3.1
func( Ie, Vm, Fr)
print Fr

## plot membrane potential trace  
plot(time, Vm)
title('Leaky Integrate-and-Fire')
ylabel('Membrane Potential (mV)')
xlabel('Time (msec)')
ylim([-70,20])
show()

Why after the func is called, the Fr is still 0?

I know it's simple but I have wasted long time on this

Thank you

newbie myself, but shouldn't you return Fr in the function? Seems like the variable is local — Subbeh
– Subbeh, Commented May 15, 2015 at 13:38
@jonrsharpe it passed the condition but the value is not incremented — Mix
– Mix, Commented May 15, 2015 at 13:40
@Subbeh thanks, the problem is the variable is local in the function — Mix
– Mix, Commented May 15, 2015 at 13:44

konart · Accepted Answer · 2015-05-15 13:59:32Z

3

You have two Fr variables in different scopes

Fr = 0

Is outside of your function, thus never changed.

Fr += 1

Is inside a function and will be incremented, but this is a different variable.

Here is the solution (one of the possible ones):

def func(Ie, Vm, Fr):
    for i, t in enumerate(time):
        if i == 0:
            Vm[i] = -70
        else: 
            Vm[i] = Vm[i-1] + (El- Vm[i-1] + Ie*Rm) / tau_m * dt
            if Vm[i] >= Vth:
                Fr += 1
                Vm[i] = El
     return Fr

Then, just do

Fr = func(Ie, Vm, Fr)

One more tip. If your Fr variable is always 0 by default you can do this:

def func(Ie, Vm, Fr=0):

when defining the function, and pass the third paramenter only when you need something different that 0.

edited May 15, 2015 at 13:59

answered May 15, 2015 at 13:40

konart

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Comments

Eloims · Accepted Answer · 2015-05-15 13:46:47Z

2

If you want to modify variable outside of the scope of the function you need to use the global keyword

my_var = True # Declare a variable on the global scope
def my_function():
     global my_var # tell the interpreter that you want to use the global "my_var"
     my_var = False # Change global my_var value

my_function() # call the function
print my_var # check result

Be advised however that it is not considered a good practice to do so.

You should try to isolate as much as you can the scopes in your code to make it more readable.

my_var = 3 # Declare a variable on the global scope
def my_function(my_var):
     return my_var + 1

my_var = my_function(my_var) # call the function and assign result to global variable
print my_var # check result

edited May 15, 2015 at 13:46

answered May 15, 2015 at 13:39

Eloims

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5 Comments

jonrsharpe Over a year ago

This is bad advice - global is rarely the correct solution.

konart Over a year ago

Using global vars is not a good practive though. I think he'd better just return the final version of Fr. In outer words Fr = func(Ie, Vm, Fr)

Mix Over a year ago

Yes, the problem is the variable is locally computed in the function, thanks for help

Eloims Over a year ago

I agree, but incrementing a variable out of the scope of a function is what the OP is asking :). This seems to be a simple script, i'm not sure it makes sense to speak about "best practices"

konart Over a year ago

Well, I think speaking about "best practices" is always makes sense, especially when it comes to a someone new to programming :)

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