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Introduction Step 1 Step 2 Step 3 Step 4 Step 5 Hints

Scientific Python Tutorial – CCN Course 2013How to code a neural network simulation

Malte J. Rasch

National Key Laboratory of Cognitive Neuroscience and LearningBeijing Normal University

July 10, 2013

Goal of tutorial

We will program a neural network simulation together.

We will practice on the way:

Writing scripts

Usage of array notation

Writing scripts

How to integrate ODEs

Writing scripts

How to plot results

Writing scripts

How to plot results

How to simulate neurons and synapses

Writing scripts

How to plot results

How to simulate neurons and synapses

How to program a quite realistic network simulation

What has to be done in principle

There are n neurons, excitatory and inhibitory, that are

inter-connected with synapses.

The network gets some input

Each neuron and each synapse follows a particular

dynamics over time.

Each neuron and each synapse follows a particular

dynamics over time.

The simulation solves the interplay of all components and

e.g. yields spiking activity of the network for given

inputs, which can be further analyzed (e.g. plotted)

We will proceed in 5 successive steps

1 Simulate a single neuron with current step input

2 Simulate a single neuron with Poisson input

3 Simulate 1000 neurons (no recurrent connections)

4 Simulate a recurrent network

5 Simulate a simple orientation column

Result plots

0 200 400 600 800 1000

Time [ms]

A single qIF neuronwith current step input

0 200 400 600 800 1000

Time [ms]

branevoltage[m

A single qIF neuronwith 100 Poisson inputs

Voltage trace

0 200 400 600 800 1000

Time [ms]

An unconnected networkof 1000 qIF neurons

0 200 400 600 800 1000

Time [ms]

An recurrent networkof 1000 qIF neurons

Which neuron model to use?

Biophysical model (i.e. Hodgkin-Huxley model)

dt= −

(Vm − VL)−∑

gi(t)(Vm − Ei) + I

Including non-linear dynamics of many channels in gi(t)

Which neuron model to use?

Biophysical model (i.e. Hodgkin-Huxley model)

dt= −

(Vm − VL)−∑

gi(t)(Vm − Ei) + I

Including non-linear dynamics of many channels in gi(t)

Mathematical simplification (Izhikevich, book chapter 8)

if v < 35 :

v = (0.04v + 5) v + 150− u − I

u = a (b v − u)

if v ≥ 35 :

v ← c

u ← u + d

With b = 0.2, c = −65, and d = 8, a = 0.02 for excitatory neurons and d = 2, a = 0.1 forinhibitory neurons.

Neuron model

regular spiking (RS) intrinsically bursting (IB) chattering (CH) fast spiking (FS)

low-threshold spiking (LTS)pa

parameter c

thalamo-cortical (TC)

-87 mV

-63 mV

thalamo-cortical (TC)

peak 30 mV

reset c

reset ddecay with rate a

sensitivity b

u(t)0 0.1

RS,IB,CH FS

LTS,TC

-65 -55 -50

FS,LTS,RZ

0.02parameter a

resonator (RZ)

v'= 0.04v 2+5v +140 - u + I u'= a(bv - u) if v = 30 mV, then v c, u u + d

Step 1: Simulate a single neuron with injected current

Exercise 1

Simulate one excitatory neuron for 1000ms and plot the resulting voltagetrace. Apply a current step (Iapp = 7pA) between time 200ms and 700ms.

Step 1: Simulate a single neuron with injected current

Exercise 1

Simulate one excitatory neuron for 1000ms and plot the resulting voltagetrace. Apply a current step (Iapp = 7pA) between time 200ms and 700ms.

Neuron model:

if v < 35 :

v = (0.04v + 5) v + 150− u − Iapp

u = a (b v − u)

if v ≥ 35 :

v ← c

u ← u + d

with d = 8, a = 0.02, b = 0.2, c = −65

0 200 400 600 800 1000

Time [ms]

A single qIF neuronwith current step input

Step 1 in detail:Open Spyder and create a new file (script) that will simulate the neuron. Import thenecessary modules (from pylab import *)

Proceed as follows:

1 Initialize parameter values (∆t = 0.5ms, a = 0.02, d = 8, · · · )

Proceed as follows:

2 Reserve memory for voltage trace v and u (of length T = 1000/∆t) and set firstelement to −70 and −14, respectively.

Proceed as follows:

3 Loop over T − 1 time steps and do for each step t

Proceed as follows:

1 set Iapp ← 7 if t∆t is between 200 and 700 (otherwise 0)

Proceed as follows:

2 if vt < 35: update element t + 1 of v and u according to

vt+1 ← vt +∆t {(0.04 vt + 5) vt − ut + 140 + Iapp}

ut+1 ← ut +∆t a (b vt − ut)

Proceed as follows:

3 if vt ≥ 35: set the variables, vt ← 35, vt+1 ← c , and ut+1 ← ut + d .

Proceed as follows:

3 if vt ≥ 35: set the variables, vt ← 35, vt+1 ← c , and ut+1 ← ut + d .

4 Plot the voltage trace v versus t

Solution to step 1 (Python)

1 from py l ab impor t ∗

3# 1) i n i t i a l i z e pa ramete r s4 tmax = 1000 .5 dt = 0 .56

7# 1 . 1 ) Neuron / Network pa r s8 a = 0.02 # RS , IB : 0 . 02 , FS : 0 . 19 b = 0 .2 # RS , IB , FS : 0 . 2

10 c = −65 # RS , FS : −65 IB : −5511 d = 8 . # RS : 8 , IB : 4 , FS : 212

13# 1 . 2 ) I npu t pa r s14 I app=1015 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt # stm t ime16

17# 2) r e s e r v e memory18T = c e i l ( tmax/ dt )19 v = z e r o s (T)20 u = z e r o s (T)21 v [ 0 ] = −70 # r e s t i n g p o t e n t i a l22 u [ 0 ] = −14 # s t eady s t a t e23

24# 3) fo r−l oop ove r t ime25 f o r t i n arange (T−1):26 # 3 . 1 ) ge t i n pu t27 i f t>t r [ 0 ] and t<t r [ 1 ] :

28 I = Iapp29 e l s e :30 I = 031

33 i f v [ t ]<35:34 # 3 . 2 ) update ODE35 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]36 v [ t +1] = v [ t ] + ( dv+I )∗ dt37 du = a ∗( b∗v [ t ]−u [ t ] )38 u [ t +1] = u [ t ] + dt ∗du39 e l s e :40 # 3 . 3 ) s p i k e !41 v [ t ] = 3542 v [ t +1] = c43 u [ t +1] = u [ t ]+d44

45# 4) p l o t v o l t a g e t r a c e46 f i g u r e ( )47 t v e c = arange ( 0 . , tmax , dt )48 p l o t ( tvec , v , ’ b ’ , l a b e l= ’ Vo l tage t r a c e ’ )49 x l a b e l ( ’ Time [ms ] ’ )50 y l a b e l ( ’Membrane v o l t a g e [mV] ’ )51 t i t l e ( ”””A s i n g l e qIF neuron52 with c u r r e n t s t e p i npu t6 ””” )53 show ( )

Synapse model

Conductance based synaptic input

A simple synaptic input model would be

Isyn =∑

wjsj(v − Ej)

where wj is the weight of the jth synapse and Ej its reversal potential (forinstance 0 mV for excitatory and −85 mV for inhibitory synapses).

Variable sj implements the dynamics of the jth synapse:

sj = −sj/τs

sj ← sj + 1, if pre-synaptic neuron spikes

Synapse model

Conductance based synaptic input

A simple synaptic input model would be

Isyn =∑

wjsj(v − Ej)

where wj is the weight of the jth synapse and Ej its reversal potential (forinstance 0 mV for excitatory and −85 mV for inhibitory synapses).

Variable sj implements the dynamics of the jth synapse:

sj = −sj/τs

sj ← sj + 1, if pre-synaptic neuron spikes

Optional: Synaptic depressionChange the update to

sj ← sj + hj , hj ← 1− (1 + (U − 1)hj)e−∆tjτd ,

with e.g. U = 0.5, τd = 500ms. ∆tj is the interval between current and previous spike ofneuron j .

Step 2: Single neuron with synaptic input

Exercise 2

Simulate the neuron model for 1000ms and plot the resulting voltage trace.Assume that 100 synapses are attached to the neuron, with each pre-synapticneuron firing with a Poisson process of rate frate = 2 Hz between time 200msand 700ms.

Step 2: Single neuron with synaptic input

Exercise 2

Simulate the neuron model for 1000ms and plot the resulting voltage trace.Assume that 100 synapses are attached to the neuron, with each pre-synapticneuron firing with a Poisson process of rate frate = 2 Hz between time 200msand 700ms.

Synaptic input model:

Isyn =∑

w inj s inj (t)(E

inj − v(t))

s inj = s inj /τs

s inj ← s inj + hj , if synapse j spikes

with hj = 1∗, τs = 10, w inj = 0.07, Ej = 0,

j = 1 . . . 100. Poisson: Input synapse j spikes if

rj(t) < frate∆t, where rj(t) ∈ [0, 1] are uniform

random numbers drawn for each step t.

0 200 400 600 800 1000

Time [ms]

branevoltage[m

A single qIF neuronwith 100 Poisson inputs

Voltage trace

Step 2 in detail:

Use the last script, save it under a new file name, and add the necessary lines.

Proceed as follows:

1 Initialize new parameter values (τs = 10, frate = 0.002ms−1)

Step 2 in detail:

Proceed as follows:

2 Reserve memory and initialize the vectors sin = (s inj ), win = (w in

j ), and E = (Ej)with nin = 100 constant elements (same values as in Step 1)

Step 2 in detail:

Proceed as follows:

3 Inside the for-loop change/add the following:

Step 2 in detail:

Proceed as follows:

1 Set pj = 1 if rj ≤ frate∆t (otherwise 0) during times of applied input. rj is anuniform random number between 0 and 1. Use array notation to set theinput for all nin input synapses. Hint

Step 2 in detail:

Proceed as follows:

1 Set pj = 1 if rj ≤ frate∆t (otherwise 0) during times of applied input. rj is anuniform random number between 0 and 1. Use array notation to set theinput for all nin input synapses. Hint

2 before the vt update: Implement the conductance dynamics s and set Iappaccording to the input. Use array notation with dot “·” product andelement-wise “⊙” product. Hint

s inj ← s inj + pj

Iapp ← win ·(

sin ⊙ Ein)

win · sin)

⊙ vt

s inj ← (1−∆t/τs) sinj

1 from py l ab impor t ∗23# 1) i n i t i a l i z e pa ramete r s4 tmax = 1000 .5 dt = 0 .567# 1 . 1 ) Neuron / Network pa r s8 a = 0.029 b = 0 .2

10 c = −6511 d = 8 .12 t a u s = 10 # decay o f s ynap s e s [ms ]1314# 1 . 2 ) I npu t pa r s15 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt16 r a t e i n = 2 # inpu t r a t e17 n i n = 100 # number o f i n p u t s18 w in = 0.07 # inpu t we i gh t s19 W in = w in∗ones ( n i n ) # v e c t o r2021# 2) r e s e r v e memory22 T = c e i l ( tmax/ dt )23 v = z e r o s (T)24 u = z e r o s (T)25 v [ 0 ] = −70 # r e s t i n g p o t e n t i a l26 u [ 0 ] = −14 # s t eady s t a t e27 s i n = z e r o s ( n i n ) # s y n a p t i c v a r i a b l e28 E i n = z e r o s ( n i n ) # r e v p o t e n t i a l29 p r a t e = dt∗ r a t e i n ∗1e−3 # abbrev3031# 3) fo r−l oop ove r t ime32 f o r t i n arange (T−1):33 # 3 . 1 ) ge t i n pu t

34 i f t>t r [ 0 ] and t<t r [ 1 ] :35 # NEW: get i n pu t Po i s son s p i k e s36 p = un i fo rm ( s i z e=n i n )<p r a t e ;37 e l s e :38 p = 0 ; # no i npu t3940 # NEW: c a l c u l a t e i n pu t c u r r e n t41 s i n = (1 − dt / t a u s )∗ s i n + p42 I = dot (W in , s i n ∗E in )43 I −= dot (W in , s i n )∗ v [ t ]4445 i f v [ t ]<35:46 # 3 . 2 ) update ODE47 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]48 v [ t +1] = v [ t ] + ( dv+I )∗ dt49 du = a∗(b∗v [ t ]−u [ t ] )50 u [ t +1] = u [ t ] + dt∗du51 e l s e :52 # 3 . 3 ) s p i k e !53 v [ t ] = 3554 v [ t +1] = c55 u [ t +1] = u [ t ]+d5657# 4) p l o t v o l t a g e t r a c e58 f i g u r e ( )59 t v e c = arange ( 0 . , tmax , dt )60 p l o t ( tvec , v , ’ b ’ , l a b e l= ’ Vo l tage t r a c e ’ )61 x l a b e l ( ’ Time [ms ] ’ )62 y l a b e l ( ’Membrane v o l t a g e [mV] ’ )63 t i t l e ( ”””A s i n g l e qIF neuron64 wi th %d Po i s son i n p u t s ””” % n i n )65 show ( )

Solution to step 2 with STP (Python)

1 from py l ab impor t ∗23# 1) i n i t i a l i z e pa ramete r s4 tmax = 1000 .5 dt = 0 .567# 1 . 1 ) Neuron / Network pa r s8 a = 0.029 b = 0 .2

10 c = −6511 d = 8 .12 t a u s = 10 # decay o f s ynap s e s [ms ]13 tau d = 500 # s y n a p t i c d e p r e s s i o n [ms ]14 s t p u = 0 .5 # STP paramete r1516# 1 . 2 ) I npu t pa r s17 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt18 r a t e i n = 10 # inpu t r a t e19 n i n = 1 # number o f i n p u t s20 w in = 0.03 # inpu t we i gh t s21 W in = w in∗ones ( n i n ) # v e c t o r2223# 2) r e s e r v e memory24 T = c e i l ( tmax/ dt )25 v = z e r o s (T)26 u = z e r o s (T)27 v [ 0 ] = −70 # r e s t i n g p o t e n t i a l28 u [ 0 ] = −14 # s t eady s t a t e29 s i n = z e r o s ( n i n ) # s y n a p t i c v a r i a b l e30 E i n = z e r o s ( n i n ) # r e v p o t e n t i a l31 p r a t e = dt∗ r a t e i n ∗1e−3 # abbrev32 h = ones ( n i n )33 l a s t s p = − i n f t y ∗ones ( n i n )3435# 3) fo r−l oop ove r t ime36 f o r t i n arange (T−1):37 # 3 . 1 ) ge t i n pu t38 i f t>t r [ 0 ] and t<t r [ 1 ] :

39 # NEW: get i n pu t Po i s son s p i k e s40 p = un i fo rm ( s i z e=n i n )<p r a t e ;4142 #update s y n a p t i c d e p r e s s i o n43 tmp = exp ( dt∗( l a s t s p [ p]− t )/ tau d )44 h [ p ] = 1 − (1+( s tp u −1)∗h [ p ] )∗ tmp45 l a s t s p [ p ] = t46 e l s e :47 p = 0 ; # no i npu t484950 # NEW: c a l c u l a t e i n pu t c u r r e n t51 s i n = (1 − dt / t a u s )∗ s i n + p∗h52 I = dot (W in , s i n ∗E in )53 I −= dot (W in , s i n )∗ v [ t ]5455 i f v [ t ]<35:56 # 3 . 2 ) update ODE57 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]58 v [ t +1] = v [ t ] + ( dv+I )∗ dt59 du = a∗(b∗v [ t ]−u [ t ] )60 u [ t +1] = u [ t ] + dt∗du61 e l s e :62 # 3 . 3 ) s p i k e !63 v [ t ] = 3564 v [ t +1] = c65 u [ t +1] = u [ t ]+d6667# 4) p l o t v o l t a g e t r a c e68 f i g u r e ( )69 t v e c = arange ( 0 . , tmax , dt )70 p l o t ( tvec , v , ’ b ’ , l a b e l= ’ Vo l tage t r a c e ’ )71 x l a b e l ( ’ Time [ms ] ’ )72 y l a b e l ( ’Membrane v o l t a g e [mV] ’ )73 t i t l e ( ”””A s i n g l e qIF neuron74 wi th %d Po i s son i n p u t s ””” % n i n )75 show ( )

Step 3: Simulate 1000 neurons (not inter-connected)

Exercise 3

Simulate 1000 neurons for 1000 ms and plot the resulting spikes. Assumethat each neuron receives (random) 10% of the 100 Poisson spike trains ofrate frate = 2 Hz between time 200 ms and 700 ms. Note that the neuronsare not yet inter-connected.

Step 3: Simulate 1000 neurons (not inter-connected)

Exercise 3

Simulate 1000 neurons for 1000 ms and plot the resulting spikes. Assumethat each neuron receives (random) 10% of the 100 Poisson spike trains ofrate frate = 2 Hz between time 200 ms and 700 ms. Note that the neuronsare not yet inter-connected.

Excitatory and inhibitory neurons:

A neuron is, with probability pinh = 0.2, a(fast-spiking) inhibitory neuron (a = 0.1,d = 2), others are (regular spiking) excitatoryneurons (a = 0.02 and d = 8).

Input weights of input synapse j to neuron i is

set to w in = 0.07 if connected (otherwise 0).

0 200 400 600 800 1000

Time [ms]

An unconnected networkof 1000 qIF neurons

Step 3 in detail:

Modify the last script (after saving it under new name).

Proceed as follows:

1 Initialize new parameter values (n = 1000)

Step 3 in detail:

Proceed as follows:

2 Initialize 2 logical vectors kinh and kexc of length n, where kinh(i) is True withprobability p = 0.2 (marking an inhibitory neuron) and False otherwise. Andkexc = ¬kinh.

Step 3 in detail:

Proceed as follows:

3 Reserve memory and initialize vi ,t , ui ,t (now being T × n matrices). Setparameter vectors a and d according to kexc and kinh.

Step 3 in detail:

Proceed as follows:

4 The weights w inij = 0.07 now form a n × nin matrix. Set 90 % random elements

to 0 to account for the connection probability.

Step 3 in detail:

Proceed as follows:

1 Same update equations (for vi ,t+1 and ui ,t+1) but use array notation.

Step 3 in detail:

Proceed as follows:

1 Same update equations (for vi ,t+1 and ui ,t+1) but use array notation.

6 Plot the spike raster. Plot black dots at {(t, i)|vit ≥ 35} for excitatory neurons i .Use red dots for inhibitory neurons.

1 from py l ab impor t ∗23# 1) i n i t i a l i z e pa ramete r s4 tmax = 1000 .5 dt = 0 .567# 1 . 1 ) Neuron / Network pa r s8 n = 1000 # number o f neurons9 p inh = 0 .2 # prob o f i nh neuron

10 inh = ( un i fo rm ( s i z e=n)<p inh ) # whether i nh .11 exc = l o g i c a l n o t ( i nh )12 a = inh . choose ( 0 . 0 2 , 0 . 1 )# exc =0.02 , i nh =0.113 b = 0 .214 c = −6515 d = inh . choose (8 , 2 ) # exc=8, i nh=216 t a u s = 101718# 1 . 2 ) I npu t pa r s19 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt20 r a t e i n = 221 n i n = 10022 w in = 0.0723 pconn in = 0 .1 # inpu t conn prob24 C = un i fo rm ( s i z e =(n , n i n ))< pconn in25 W in = C . choose (0 , w in ) # mat r i x2627# 2) r e s e r v e memory28 T = c e i l ( tmax/ dt )29 v = z e r o s ( (T, n ) ) # now mat r i x30 u = z e r o s ( (T, n ) ) # now mat r i x31 v [ 0 ] = −70 # s e t 1 s t row32 u [ 0 ] = −1433 s i n = z e r o s ( n i n )34 E i n = z e r o s ( n i n )35 p r a t e = dt∗ r a t e i n ∗1e−33637# 3) fo r−l oop ove r t ime38 f o r t i n arange (T−1):39 # 3 . 1 ) ge t i n pu t40 i f t>t r [ 0 ] and t<t r [ 1 ] :

41 p = un i fo rm ( s i z e=n i n )<p r a t e ;42 e l s e :43 p = 0 ;4445 s i n = (1 − dt / t a u s )∗ s i n + p46 I = W in . dot ( s i n ∗E in )47 I −= W in . dot ( s i n )∗ v [ t ]4849 # NEW: hand l e a l l neu rons50 f i r e d = v [ t ]>=355152 # 3 . 2 ) update ODE, s imp l y update a l l53 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]54 v [ t +1] = v [ t ] + ( dv+I )∗ dt55 du = a∗(b∗v [ t ]−u [ t ] )56 u [ t +1] = u [ t ] + dt∗du5758 # 3 . 3 ) s p i k e s !59 v [ t ] [ f i r e d ] = 3560 v [ t +1] [ f i r e d ] = c61 u [ t +1] [ f i r e d ] = u [ t ] [ f i r e d ]+d [ f i r e d ]6263# 4) p l o t t i n g64# NEW: get s p i k e s and p l o t65 tspk , nspk = nonzero ( v==35)66 i d x i = in1d ( nspk , nonze ro ( i nh ) [ 0 ] ) # f i n d i nh67 i d x e = l o g i c a l n o t ( i d x i ) # a l l o t h e r s a r e exc6869 f i g u r e ( )70 p l o t ( t spk [ i d x e ]∗ dt , nspk [ i d x e ] , ’ k . ’ ,71 l a b e l= ’ Exc . ’ , ma r k e r s i z e =2)72 p l o t ( t spk [ i d x i ]∗ dt , nspk [ i d x i ] , ’ r . ’ ,73 l a b e l= ’ Inh . ’ , ma r k e r s i z e =2)74 x l a b e l ( ’ Time [ms ] ’ )75 y l a b e l ( ’ Neuron number [\#] ’ )76 x l im ( ( 0 , tmax ) )77 t i t l e ( ”””An unconnected network78 o f %d qIF neurons ””” % n)79 l eg end ( l o c=’ upper r i g h t ’ )80 show ( )

Step 4: Simulate recurrent network

Exercise 4

Simulate 1000 neurons as before but with added recurrent connections.

Step 4: Simulate recurrent network

Exercise 4

Simulate 1000 neurons as before but with added recurrent connections.

Recurrent synaptic activations

A neuron i is sparsely connected to a neuron j

(with probability pconn = 0.1). Thus neuron i

receives an additional current I syni of the form:

Isyni =

wijsj(t) (Ej − vi (t))

Weights are Gamma distributed(wavg = 0.005 and gsc = 0.002). Set theinhibitory to excitatory connections twice asstrong on average.

0 200 400 600 800 1000

Time [ms]

Step 4 in detail:

Proceed as follows:

1 Initialize and allocate memory for the new variables (s = (sj), Ej). Set Ej = −85if j is an inhibitory neuron (otherwise 0).

Step 4 in detail:

Proceed as follows:

2 Reserve memory and initialize weights W = (wij) to zero. Randomly choose 10%of the matrix elements.

Step 4 in detail:

Proceed as follows:

3 Set the chosen weight matrix elements to values drawn from a Gammadistribution of scale gsc = 0.002 and shape gsh =

gsc, with wavg = 0.005. Hint

Step 4 in detail:

Proceed as follows:

4 Make the weight matrix “sparse” to speed up computations. Hint

Step 4 in detail:

Proceed as follows:

5 Scale weights from inh. to exc. neurons by the factor of 2. Hint

Step 4 in detail:

Proceed as follows:

5 Scale weights from inh. to exc. neurons by the factor of 2. Hint

1 add the equations for recurrent synaptic dynamics sj and add Isyn to the totalapplied current.

sj ← sj + 1, if vj(t − 1) ≥ 35

Isyn ← W · (s⊙ E)− (W · s)⊙ v

sj ← (1−∆t/τs) sj

Solution to step 41 from py l ab impor t ∗2 from s c i p y . s p a r s e impor t c s r m a t r i x34 # 1) i n i t i a l i z e pa ramete r s5 tmax = 1000 .6 dt = 0 .578# 1 . 1 ) Neuron / Network pa r s9 n = 1000

10 p inh = 0 .211 inh = ( un i fo rm ( s i z e=n)<p inh )12 exc = l o g i c a l n o t ( i nh )13 a = inh . choose ( 0 . 0 2 , 0 . 1 )14 b = 0 .215 c = −6516 d = inh . choose (8 , 2 )17 t a u s = 101819# NEW r e c u r r e n t paramete r20 w = 0.005 # ave rage r e c u r r e n t we ight21 pconn = 0 .1 # r e c u r r e n t conne c t i on prob22 s c a l e E I = 2 # s c a l e I−>E23 g s c = 0.002 # s c a l e o f gamma24 E = inh . choose (0 ,−85)25# NEW: make we ight mat r i x26W = ze r o s ( ( n , n ) )27 C = un i fo rm ( s i z e =(n , n ) )28 i d x = nonzero (C<pconn ) # sp a r s e c o n n e c t i v i t y29W[ i d x ] = gamma(w/ g sc , s c a l e=g sc , s i z e=i d x [ 0 ] . s i z e )30W[ i x ( exc , i nh ) ] ∗= s c a l e E I #submat i n d e x i n g31W = c s r m a t r i x (W) # make row s p a r s e3233# 1 . 2 ) I npu t pa r s34 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt35 r a t e i n = 236 n i n = 10037 w in = 0.0738 pconn in = 0 .139 C = un i fo rm ( s i z e =(n , n i n ))< pconn in40 W in = C . choose (0 , w in )4142# 2) r e s e r v e memory

43 T = c e i l ( tmax/ dt )44 v = z e r o s ( (T, n ) )45 u = z e r o s ( (T, n ) )46 v [ 0 ] = −7047 u [ 0 ] = −1448 s i n = z e r o s ( n i n )49 E i n = z e r o s ( n i n )50 p r a t e = dt∗ r a t e i n ∗1e−351 s = z e r o s ( n ) # r e c s ynap s e s5253# 3) fo r−l oop ove r t ime54 f o r t i n arange (T−1):55 # 3 . 1 ) ge t i n pu t56 i f t>t r [ 0 ] and t<t r [ 1 ] :57 p = un i fo rm ( s i z e=n i n )<p r a t e ;58 e l s e :59 p = 0 ;6061 s i n = (1 − dt / t a u s )∗ s i n + p62 I = W in . dot ( s i n ∗E in )63 I −= W in . dot ( s i n )∗ v [ t ]6465 f i r e d = v [ t ]>=356667 # NEW: r e c u r r e n t i n pu t68 s = (1 − dt / t a u s )∗ s + f i r e d69 I s y n = W. dot ( s∗E) − W. dot ( s )∗ v [ t ]70 I += I s y n # add to i npu t v e c t o r7172 # 3 . 2 ) update ODE73 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]74 v [ t +1] = v [ t ] + ( dv+I )∗ dt75 du = a∗(b∗v [ t ]−u [ t ] )76 u [ t +1] = u [ t ] + dt∗du7778 # 3 . 3 ) s p i k e s !79 v [ t ] [ f i r e d ] = 3580 v [ t +1] [ f i r e d ] = c81 u [ t +1] [ f i r e d ] = u [ t ] [ f i r e d ]+d [ f i r e d ]8283# 4) p l o t t i n g84 tspk , nspk = nonzero ( v==35)

Step 5: Simulate an orientation column

Exercise 5

Restructure the connection matrix and the input to simulate an orientationcolumn. That is all E-E neurons only connect to neighboring neurons and thenetwork resembles a 1D ring.

Step 5: Simulate an orientation column

Exercise 5

Restructure the connection matrix and the input to simulate an orientationcolumn. That is all E-E neurons only connect to neighboring neurons and thenetwork resembles a 1D ring.

Ring structure

A neuron i is still sparsely connected to aneuron j but now with probability pconn = 0.4.However, if all neurons are arranged on a ringfrom 0 to 2π exc-to-exc connections are onlypossible if two neurons are nearer than π/4.Input is only delivered to a half of neurons(e.g. from 0 to pi). Use the same inputconnection probability as before.

0 200 400 600 800 1000

Time [ms]

Step 5 in detail:

Proceed as follows:

1 Set indexes of the weight matrix to zero, which belong to exc-exc connectionsfurther apart that π/4. Hint: One can use scipy.linalg.circulant

Step 5 in detail:

Proceed as follows:

1 Set indexes of the weight matrix to zero, which belong to exc-exc connectionsfurther apart that π/4. Hint: One can use scipy.linalg.circulant

2 Change the input so that only half (e.g. from 0 to pi) of the neurons receiveinput (again with probability 0.2). neurons

Solution to step 51 from py l ab impor t ∗2 from s c i p y . s p a r s e impor t c s r m a t r i x3 from s c i p y . l i n a l g impor t c i r c u l a n t4 # 1) i n i t i a l i z e pa ramete r s5 tmax = 1000 .6 dt = 0 .578# 1 . 1 ) Neuron / Network pa r s9 n = 1000

10 p inh = 0 .211 inh = ( un i fo rm ( s i z e=n)<p inh )12 exc = l o g i c a l n o t ( i nh )13 a = inh . choose ( 0 . 0 2 , 0 . 1 )14 b = 0 .215 c = −6516 d = inh . choose (8 , 2 )17 t a u s = 101819 width = p i /4 # ha l f−width o f the o r i e n t a t i o n tun i ng20 w = 0.00521 pconn = 0 .4 # s e t a b i t h i g h e r22 s c a l e E I = 223 g s c = 0.00224 E = inh . choose (0 ,−85)25W = ze r o s ( ( n , n ) )26 C = un i fo rm ( s i z e =(n , n ) )27 i d x = nonzero (C<pconn )28W[ i d x ] = gamma(w/ g sc , s c a l e=g sc , s i z e=i d x [ 0 ] . s i z e )29W[ i x ( exc , i nh ) ] ∗= s c a l e E I30 th e t a = l i n s p a c e (0 ,2∗ pi , n ) # NEW31 R = c i r c u l a n t ( cos ( t h e t a ))> cos ( width ) #NEW32W[ : , exc ] = where (R [ : , exc ] ,W[ : , exc ] , 0 ) # NEW33W = c s r m a t r i x (W)3435# 1 . 2 ) I npu t pa r s36 t r=a r r a y ( [ 2 0 0 . , 7 0 0 ] ) / dt37 r a t e i n = 238 i nw i d t h = p i /239 w in = 0.0740 pconn in = 0 .241 n i n = 10042 C = un i fo rm ( s i z e =(n , n i n ))< pconn in

43 W in = C . choose (0 , w in )44 W in [ n / 2 : , : ] = 0 # NEW4546# 2) r e s e r v e memory47 T = c e i l ( tmax/ dt )48 v = z e r o s ( (T, n ) )49 u = z e r o s ( (T, n ) )50 v [ 0 ] = −7051 u [ 0 ] = −1452 s i n = z e r o s ( n i n )53 E i n = z e r o s ( n i n )54 p r a t e = dt∗ r a t e i n ∗1e−355 s = z e r o s ( n ) # r e c s ynap s e s5657# 3) fo r−l oop ove r t ime58 f o r t i n arange (T−1):59 # 3 . 1 ) ge t i n pu t60 i f t>t r [ 0 ] and t<t r [ 1 ] :61 p = un i fo rm ( s i z e=n i n )<p r a t e ;62 e l s e :63 p = 0 ;6465 s i n = (1 − dt / t a u s )∗ s i n + p66 I = W in . dot ( s i n ∗E in )67 I −= W in . dot ( s i n )∗ v [ t ]6869 f i r e d = v [ t ]>=357071 # NEW: r e c u r r e n t i n pu t72 s = (1 − dt / t a u s )∗ s + f i r e d73 I s y n = W. dot ( s∗E) − W. dot ( s )∗ v [ t ]74 I += I s y n # add to i npu t v e c t o r7576 # 3 . 2 ) update ODE77 dv = (0 .04∗ v [ t ]+5)∗ v [ t ]+140−u [ t ]78 v [ t +1] = v [ t ] + ( dv+I )∗ dt79 du = a∗(b∗v [ t ]−u [ t ] )80 u [ t +1] = u [ t ] + dt∗du8182 # 3 . 3 ) s p i k e s !83 v [ t ] [ f i r e d ] = 3584 v [ t +1] [ f i r e d ] = c

Congratulation !

You have just coded and simulated a quite realisticnetwork model !

Hint for generating random number

Use the uniform function to generate arrays of random numbersbetween 0 and 1.

1 from numpy . random impor t un i fo rm2 n i n = 1003 r = un i fo rm ( s i z e=n i n )

To set indexes i of a vector v to a with a probability p and otherwise tob one can use the method choose

1 r = un i fo rm ( s i z e=n i n )2 v = ( r<p ) . choose (b , a )

Hint for generating gamma distributed random number

Use the numpy.random.gamma function to generate arrays of randomnumbers which are gamma distributed.

1 from numpy . random impor t gamma2

3 g shape , g s c a l e , n = 0 .003 , 2 , 10004 r = gamma( g shape , g s c a l e , n )

Hint for making sparse matrices

There are several forms of sparse matrices in the modulescipy.sparse. The one which is interesting for our purposes is the“row-wise” sparse matrix (see the documentation of scipy.sparse formore information).

1 from py l ab impor t ∗2 from s c i p y . s p a r s e impor t c s r m a t r i x3

4R = un i fo rm ( s i z e =(100 ,100)) # example mat r i x5W = where (R<0.1 ,R , 0 ) # most l y 06W2 = c s r m a t r i x (W) # make s p a r s e mat r i x7

8 v = un i fo rm ( s i z e =100) # example v e c t o r9 x = W2. dot ( v ) # mat r i x dot p roduc t

Hint for submatrix indexing

numpy.array provides a shortcut for (MatLab-like) submatrixindexing. Assume one has a matrix W and wanted to add 1 to add 1 toa selection of rows and columns. One could use the convenient functionix and write

1 from py l ab impor t ∗2

3W = un i fo rm ( s i z e =(10 ,10)) # example mat r i x4 i r ow = un i fo rm ( s i z e =10)<0.5 # s e l e c t some rows5 i c o l = un i fo rm ( s i z e =10)<0.5 # s e l e c t some c o l s6

7W( i x ( i r ow , i c o l ) ) += 1 # add 1 to the e l emen t s

Hint for using dot product

Use the dot method of an numpy.array to compute the dot product.Caution: The operator * yields an element-wise multiplication!

1 from py l ab impor t ∗2 a = a r r a y ( [ 1 . , 2 , 3 , 4 ] )3 b = a r r a y ( [ 1 . , 1 , 5 , 5 ] )4 c = a∗b # element−wi s e !5 c . s i z e6 47 d = a . dot ( b ) # s c a l a r p roduc t8 d . s i z e9 1

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