CODEWITHSUNDEEP
pythonarraysnumpywhere-clause
Nihilum
Nihilum

Reputation: 711

Error with finding the index of value closest to 0

I have an issue with my code. I am trying to find the x value for which my y is equal (or close) to 0, for the first time.

The following code computes the values of x and y for different parameters. What I don't understand is that when I try the code line by line, I obtain a min-value for my val array, but using np.where doesn't work as it can't find this value it just obtained.. I thought it was maybe related to how the min_value is stored and that it somehow rounds it up, making it not equal to what it previously was. I tried using `idx = np.where(np.isclose(val,0)) instead but I couldn't make it work. How can I solve this problem ?

D = np.logspace(21,27,7)
x = np.logspace(0,7,500)
V0 = 1.17e+13
alpha = ((4*D)/((3300-1000)*9.81))**(1/4)


fig, axs = plt.subplots(2,4, figsize=(15, 6), facecolor='w', edgecolor='k')
fig.subplots_adjust(hspace = .5, wspace=.001)

axs = axs.ravel()

for i in range(0,len(D)):
  val = np.zeros((len(x)))

  val = -(V0*(alpha[i]**3))/(8*D[i]) * np.exp(-x/alpha[i]) * (np.cos(x/alpha[i]) + np.sin(x/alpha[i]))

  # Find the index for which val crosses the x axis for the first time
  min_val = np.min(np.abs(val))
  idx = np.where(val == min_val)
  idx_first = idx[0][0]
  pos = x[idx_first]

Upvotes: 0

Views: 69

Answers (1)

Joe Todd
Joe Todd

Reputation: 897

It's because your minimum value is negative, so when you take

min_val = np.min(np.abs(val))

min_val isn't a member of your array. But it is a member of np.abs(val) so I'd recommend:

idx = np.where(np.abs(val) == min_val)

Upvotes: 1

Related Questions