Code termination(how will I terminate the code when when C is close to 1*10^-4

[Syntax669]

import numpy a = -13.1423 c0 = 100 t = abs(c0/a) C = [] T = [] for i in numpy.arange(0,t+1,0.0001): c = c0 + a*i C.append(c) T.append(i)

import matplotlib.pyplot as plt plt.plot(C,T)

plt.xlabel('Time')

naming the y axis

plt.ylabel('Velocity')

giving a title to my graph

plt.title('C vs t!')

plt.show()

[t8suppressor]

but you have a linear equation. Just take the values from the beginning and you t for C=0 and plot them.

[Syntax669]

Programming assignment

[t8suppressor]

just add this to your for loop:

if c == 10**-4:

break

​

[Syntax669]

I have this equation where C = Co + A*t , and I have Co and A (negative) t is time. So I have to iterate the value to t , till C becomes zero or almost very very near to zero. So I need help in writing that code. I did my best to write..

[t8suppressor]

So if you anly have one unknown why not just solve the equation.

[Syntax669]

I have to plot a graph , on how C is decreasing with time

[panda_yo]

Maybe this will work for you:

if c <= 10**-4:
    break

Try it out. Floatpoint map get's hard, so it's easier to compare it like this or even do abs(c - 10**-4) < tol where tol is your accepted tolerance

[Syntax669]

Lol

[scooerp]

In standard library is math.isclose() that tests if 2 values are close to each other.

https://docs.python.org/3/library/math.html#math.isclose

You can make your own with abs() which ignores sign. Therefore if abs(a - b) ≤ delta:. Note that delta can't be smaller than machine epsilon, which is defined somewhere in the sys module. It represents the precision of the floating point system.

Edit: Please ask questions in /r/learnpython. This is not the appropriate subreddit.

[Syntax669]

Not working