WordPress Custom Post Status not Public but Viewable Through Link: Privately Published Posts

I have been working on my site to bring up private posts on my site. I built a custom post status for this purpose.

My goal is to exclude posts belonging to that post status from search, from Blog Page, from WordPress Queries, RSS Feed etc. Basically there should be no mentions of it anywhere on site.

Till here it seems simple, but the problem is that I WANT people to be able to view the post via direct link.

Basically I want it to be not mentioned throughout my site, but if someone visits it from direct link, they can view it.

function custom_post_status(){  register_post_status( 'privatised', array( 'label' => _x( 'Privately Published', array('post', 'download')), 'public' => false,  'exclude_from_search' => true, 'show_in_admin_all_list' => true,  'show_in_admin_status_list' => true,  'label_count' => _n_noop( 'Privately Published <span class="count">(%s)</span>', 'Aggregated <span class="count">(%s)</span>' ),  ) );  }  add_action( 'init', 'custom_post_status' );  

Here if you see in the Array, the first option is Public, and I set it to false, which basically ensures that my post is not there on my Blog Page, or search, or Taxonomy Page, or RSS FEED etc. Etc. But if I open the same post via it’s direct url, then it gives a 404, because not found…

Please help, me have my post visible only through the direct link of it.

As a sidenote, Is it possible that I can have it public, and then exclude it from Taxonomy and all, like I can do with exclude from search…

What’s the lowest level at which a PC can break through a hewn stone wall in 1 round?

This question asks how PCs can break a stone wall. Aside from the usage of disintigrate to immediately destroy a wall, it seems that a wall can be destroyed by reducing it’s HP to 0. Walls have a hardness which reduces the damage done to them.

What is the lowest level at which a player character can destroy a wall in 1 round? Magical and mundane means are both in scope, and any item which the character has access to. Limit items to the character’s level or lower.

Find how by which actions users move through my website

I have a very simple e-commerce.

As you can see in the image, people move from the product page to the checkout page and then in most cases back to the product page.

Might be that people just click on the back button, but maybe there is some other way (maybe some error?) that makes them go back to the product page. I would love to know what people do on the checkout page to go to the product page so I can think of what to do to avoid this and make the confirm their order.

Google Analytics only shows me the flow, but not the actions taken.

Is there any way or tool that can help me?

enter image description here

Does a successful Tumble Through trigger reactions?

The text of tumble through says:

You Stride up to your Speed. During this movement, you can try to move through the space of one enemy. Attempt an Acrobatics check against the enemy’s Reflex DC as soon as you try to enter its space. You can Tumble Through using Climb, Fly, Swim, or another action instead of Stride in the appropriate environment.

This doesn’t mention that it won’t trigger reactions such as attacks of opportunity. However in Failure it states:

Your movement ends, and you trigger reactions as if you had moved out of the square you started in.

Which implies that success won’t trigger those reactions.

After running through my algorithm fine, my array returns as all zeros

I’ve spent a long time finally getting my algorithm in Python to invert Laplace transforms and live plot them to work, but unfortunately, although it seems to run fine within the function, it continuously just plots zeros!

Blue = estimation(shouldn't be zero!), Red = actual

I’ve wasted so much time getting this thing to work, and now I have no clue what could be wrong. So forgive me but I think I’ll have to post the whole thing.

The equation this is based on is a series-integral and it iterates until the series converges. However, it converges super slowly so I used Aitken’s delta-squared method to speed the whole thing up. The goal is to populate array y with each Aitken’s iteration from 0.000001 to 10 and then compare the estimation’ integral with the previous iteration’s if a random point in the former is close enough to the latter. This loops until the integrals have a difference of less than 0.01 or a set amount of iterations is reached (~30000).

This happens for 6 equations simultaneously using parallel processing. The graph live plots all 6 (in blue) compared to each of the actual equations they’re converging towards (in red) using matplotlib.

For debugging, I basically tried putting a bunch of print statements everywhere in the f_u function to see what was going on. Everything looked like it was going fine. But in the __main__ function and plot functions when I did the same, I was getting all zeros for the same array. I thought it was a local vs. global issue so I put global y in the plot and f_u, with no luck.

I know this isn’t doing this due to the parallelization; I tried without it with no luck. Basically, I suggest looking at what y does between the f_u function and the __main__ function, thats where the relationship with the y arrays breaks down I think, I just don’t know what to do to fix it.

To execute this, you’re gonna need all the packages in the import statement, including numba. You can remove the @jit decorations on the Laplace functions if you don’t want numba, but it will be sloooooow.

Here it is:

import matplotlib.pyplot as plt from matplotlib import style from matplotlib import animation import numpy as np from scipy.integrate import quad, simps from scipy.special import jv import math import random import datetime import multiprocessing as mp from numba import jit  start = datetime.datetime.now()     # Record time start  # Initialize plot style.use('fivethirtyeight')  fig, axes = plt.subplots(3, 2) ((ax1, ax2), (ax3, ax4), (ax5, ax6)) = axes  plt.autoscale(enable=True)  for ax in axes.flat:     ax.label_outer()  # Initialize variables gamma = np.array([np.nan, 1, 1, 0, 1.5, 2, 1])    # Bromwich contour parameter for each equation num = 6     # Number of equations to be processed in total n = 0   # Repetition constant for equation 3, set to either zero or 1 (diverges at 1) max_count = 1000   # Number of data points in x and y epsilon = 10 ** -40     # Minimum number to divide by in Aitken's iteration err = 10 ** -4    # Maximum difference in y to trigger comparison in closing() max_err = 0.01    # Maximum difference in integrals computed in closing() to end iteration process  x_0 = 0     # Set first integral to 0 for now result = np.zeros(7)    # Vector to store Aitken's iteration for closing() comparison previous = np.zeros(7)     # Vector to store iteration before Aitken's for closing() comparison end = np.zeros(7)     # Vector to store end conditions -- running = 0, ended = 1  u, step = np.linspace(0.000001, 10, max_count, retstep=True)    # Initializing x values  y = np.empty([num, 4, max_count])     # Initializing y values -- form is y[equation number][iteration in Aitken's][x value] y_denom = np.empty([num, max_count])    # Initializing array to store Aitken's denominator values for each equation x_iter = np.empty([num, max_count])    # Initializing array to store Aitken's iteration as starting point for next iteration  def printer(num, k):    # Prints the equation number, iteration and y value at 10     if num == 1:         print(str(num) + "    " + str(k) + "    " + str(y[0][3][max_count-1]) + "\n")     elif num == 2:         print(str(num) + "    " + str(k) + "    " + str(y[1][3][max_count-1]) + "\n")     elif num == 3:         print(str(num) + "    " + str(k) + "    " + str(y[2][3][max_count-1]) + "\n")     elif num == 4:         print(str(num) + "    " + str(k) + "    " + str(y[3][3][max_count-1]) + "\n")     elif num == 5:         print(str(num) + "    " + str(k) + "    " + str(y[4][3][max_count-1]) + "\n")     elif num == 6:         print(str(num) + "    " + str(k) + "    " + str(y[5][3][max_count-1]) + "\n")  def plot1():    # Plots estimation 1 (blue) and actual equation 1 (red)     global y     ax1.clear()     line1, = ax1.plot(u, y[0][3][0:max_count])     ax1.plot(u, np.sin(u), 'tab:red')     return line1,  def plot2():    # Plots estimation 2 (blue) and actual equation 2 (red)     global y     ax2.clear()     line2, = ax2.plot(u, y[1][3][0:max_count])     ax2.plot(u, np.heaviside(u, 1), 'tab:red')     return line2,  def plot3():    # Plots estimation 3 (blue) and actual equation 3 (red)     global y     ax3.clear()     line3, = ax3.plot(u, y[2][3][0:max_count])     ax3.plot(u, np.cos(2 * np.sqrt(u)) / np.sqrt(np.pi * u), 'tab:red')     return line3,  def plot4():    # Plots estimation 4 (blue) and actual equation 4 (red)     global y     ax4.clear()     line4, = ax4.plot(u, y[3][3][0:max_count])     ax4.plot(u, np.log(u), 'tab:red')     return line4,  def plot5():    # Plots estimation 5 (blue) and actual equation 5 (red)     global y     ax5.clear()     line5, = ax5.plot(u, y[4][3][0:max_count])     ax5.plot(u, np.heaviside(u - 1, 1) - np.heaviside(u - 2, 1), 'tab:red')     return line5,  def plot6():    # Plots estimation 6 (blue) and actual equation 6 (red)     global y     ax6.clear()     line6, = ax6.plot(u, y[5][3][0:max_count])     ax6.plot(u, jv(0, u), 'tab:red')     return line6,  def animate(i):     # Plots all equations, used to live-update graph     ln1, = plot1()     ln2, = plot2()     ln3, = plot3()     ln4, = plot4()     ln5, = plot5()     ln6, = plot6()     return ln1, ln2, ln3, ln4, ln5, ln6,  @jit(nopython=True, cache=True) def f_p1(omega, u, k):      # Equation 1, Laplace form     b = (omega + k * np.pi) / u     a = gamma[1]      f1 = a * (1 / (a ** 2 + (b + 1) ** 2) + 1 / (a ** 2 + (b - 1) ** 2)) * np.cos(omega)     return f1  @jit(nopython=True, cache=True) def f_p2(omega, u, k):      # Equation 2, Laplace form     b = (omega + k * np.pi) / u     a = gamma[2]      f2 = a / (a ** 2 + b ** 2) * np.cos(omega)     return f2  @jit(nopython=True, cache=True) def f_p3(omega, u, k):      # Equation 3, Laplace form     b = (omega + k * np.pi) / u     a = gamma[3]      f3 = np.e ** (-a / (a ** 2 + b ** 2)) * (a ** 2 + b ** 2) ** (1 / 4) * np.cos(         (math.atan2(b, a) + 2 * n * np.pi) / 2) * np.cos(b / (a ** 2 + b ** 2)) / (                 np.sqrt(a ** 2 + b ** 2) * (np.cos((math.atan2(b, a) + 2 * n * np.pi) / 2) ** 2 + np.sin(                     (math.atan2(b, a) + 2 * n * np.pi) / 2))) * np.cos(omega)     return f3  @jit(nopython=True, cache=True) def f_p4(omega, u, k):      # Equation 4, Laplace form     b = (omega + k * np.pi) / u     a = gamma[4]      f4 = -(a * np.log(a ** 2 + b ** 2) + a * np.e) / (a ** 2 + b ** 2) * np.cos(omega)     return f4  @jit(nopython=True, cache=True) def f_p5(omega, u, k):      # Equation 5, Laplace form     b = (omega + k * np.pi) / u     a = gamma[5]      f5 = a * (np.e ** (-a) * np.cos(b) - np.e ** (-2 * a) * np.cos(2 * b)) / (a ** 2 + b ** 2) * np.cos(omega)     return f5  @jit(nopython=True, cache=True) def f_p6(omega, u, k):      # Equation 6, Laplace form     b = (omega + k * np.pi) / u     a = gamma[6]      f6 = np.sqrt((a ** 2 - b ** 2 + 1) ** 2 + (2 * a * b) ** 2) * np.cos(         (math.atan2(2 * a * b, a ** 2 - b ** 2 + 1) + 2 * n * np.pi) / 2) / (             ((a ** 2 - b ** 2 + 1) ** 2 + (2 * a * b) ** 2) * (                 np.cos((math.atan2(2 * a * b, a ** 2 - b ** 2 + 1) + 2 * n * np.pi) / 2) ** 2 + np.sin(                     (math.atan2(2 * a * b, a ** 2 - b ** 2 + 1) + 2 * n * np.pi) / 2) ** 2)) * np.cos(omega)     return f6  def closing(result, previous, num, k):    # Checks for convergence. If so, adds each term to f_0 and ends process     if abs(result - previous) < max_err:         diff = abs(result - previous)          print("Equation " + str(num) + " converges to " + str(diff) + " with " + str(k) + " iterations." + "\n")          if num == 1:             for i in range(max_count):                 y[0][3][i] += f_0(num, i)             end[1] += 1             return 1         elif num == 2:             for i in range(max_count):                 y[1][3][i] += f_0(num, i)             end[2] += 1             return 1         elif num == 3:             for i in range(max_count):                 y[2][3][i] += f_0(num, i)             end[3] += 1             return 1         elif num == 4:             for i in range(max_count):                 y[3][3][i] += f_0(num, i)             end[4] += 1             return 1         elif num == 5:             for i in range(max_count):                 y[4][3][i] += f_0(num, i)             end[5] += 1             return 1         elif num == 6:             for i in range(max_count):                 y[5][3][i] += f_0(num, i)             end[6] += 1             return 1     else:         print("Equation " + str(num) + " has not yet converged.\n")         return 0  def f_0(num, i):     # Initial integral in the series. Added at end after closing() convergence test     if num == 1:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p1, 0, np.pi / 2, args=(u[i], 0))[0]     elif num == 2:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p2, 0, np.pi / 2, args=(u[i], 0))[0]     elif num == 3:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p3, 0, np.pi / 2, args=(u[i], 0))[0]     elif num == 4:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p4, 0, np.pi / 2, args=(u[i], 0))[0]     elif num == 5:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p5, 0, np.pi / 2, args=(u[i], 0))[0]     elif num == 6:         x_0 = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * quad(             f_p6, 0, np.pi / 2, args=(u[i], 0))[0]      return x_0  def f_u(u, num):    # Main algorithm. Here, the summation integrals iterate     global y, result, previous    # Forces retrieval of from global of these arrays? (attempt to debug)     k = 1   # Number of iterations      for i in range(max_count):  # First overall series iteration (initial 1st)         if num == 1:             y[0][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p1, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]         elif num == 2:             y[1][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p2, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]         elif num == 3:             y[2][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p3, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]         elif num == 4:             y[3][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p4, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]         elif num == 5:             y[4][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p5, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]         elif num == 6:             y[5][0][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                 f_p6, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]      for j in range(10000):   # Iteration loop. Goes until closing() or k = ~30000 for each equation         if j > 0:             for i in range(max_count):  # Sets the first iteration (out of 3 for Aitken's) equal to the previous Aitken's or the first iteration if j = 0 (1st)                 if num == 1:                     y[0][0][i] = x_iter[0][i]                 elif num == 2:                     y[1][0][i] = x_iter[1][i]                 elif num == 3:                     y[2][0][i] = x_iter[2][i]                 elif num == 4:                     y[3][0][i] = x_iter[3][i]                 elif num == 5:                     y[4][0][i] = x_iter[4][i]                 elif num == 6:                     y[5][0][i] = x_iter[5][i]             else:                 k += 1          for i in range(max_count):  # Second iteration for Aitken's (2nd)             if num == 1:                 y[0][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p1, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 2:                 y[1][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p2, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 3:                 y[2][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p3, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 4:                 y[3][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p4, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 5:                 y[4][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p5, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 6:                 y[5][1][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p6, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]          k += 1          for i in range(max_count):  # Third iteration for Aitken's (3rd)             if num == 1:                 y[0][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p1, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 2:                 y[1][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p2, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 3:                 y[2][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p3, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 4:                 y[3][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p4, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 5:                 y[4][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p5, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]             elif num == 6:                 y[5][2][i] = 2 * np.e ** (gamma[num] * u[i]) / (np.pi * u[i]) * (-1) ** k * quad(                     f_p6, -np.pi / 2, np.pi / 2, args=(u[i], k))[0]          k += 1          for i in range(max_count):  # Aitken's delta-squared method iteration using the previous 3 (4th)             if num == 1:                 y_denom[0][i] = (y[0][2][i] - y[0][1][i]) - (y[0][1][i] - y[0][0][i])                  if abs(y_denom[0][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[0][3][i] = y[0][2][i] - ((y[0][2][i] - y[0][1][i]) ** 2) / y_denom[0][i]             elif num == 2:                 y_denom[1][i] = (y[1][2][i] - y[1][1][i]) - (y[1][1][i] - y[1][0][i])                  if abs(y_denom[1][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[1][3][i] = y[1][2][i] - ((y[1][2][i] - y[1][1][i]) ** 2) / y_denom[1][i]             elif num == 3:                 y_denom[2][i] = (y[2][2][i] - y[2][1][i]) - (y[2][1][i] - y[2][0][i])                  if abs(y_denom[2][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[2][3][i] = y[2][2][i] - ((y[2][2][i] - y[2][1][i]) ** 2) / y_denom[2][i]             elif num == 4:                 y_denom[3][i] = (y[3][2][i] - y[3][1][i]) - (y[3][1][i] - y[3][0][i])                  if abs(y_denom[3][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[3][3][i] = y[3][2][i] - ((y[3][2][i] - y[3][1][i]) ** 2) / y_denom[3][i]             elif num == 5:                 y_denom[4][i] = (y[4][2][i] - y[4][1][i]) - (y[4][1][i] - y[4][0][i])                  if abs(y_denom[4][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[4][3][i] = y[4][2][i] - ((y[4][2][i] - y[4][1][i]) ** 2) / y_denom[4][i]             elif num == 6:                 y_denom[5][i] = (y[5][2][i] - y[5][1][i]) - (y[5][1][i] - y[5][0][i])                  if abs(y_denom[5][i]) < epsilon:                     print("Denominator too small to calculate. Exiting...\n")                     return 9                  y[5][3][i] = y[5][2][i] - ((y[5][2][i] - y[5][1][i]) ** 2) / y_denom[5][i]          k += 1         printer(num, k)          rand = random.randrange(0, max_count)   # Setting random number for x value          # Comparison between y at random x in Aitken's vs the previous iteration. If close enough, triggers closing()         if num == 1:             if abs(y[0][3][rand] - y[0][2][rand]) < err:                 result[num] = simps(y[0][3][0:max_count], u, dx=step)   #Integrates Aitken's iteration (4th)                 previous[num] = simps(y[0][2][0:max_count], u, dx=step)    #Integrates integration before Aitken's (3rd)                 c1 = closing(result[num], previous[num], num, k)                 if c1 == 1:                     break         elif num == 2:             if abs(y[1][3][rand] - y[1][2][rand]) < err:                 result[num] = simps(y[1][3][0:max_count], u, dx=step)                 previous[num] = simps(y[1][2][0:max_count], u, dx=step)                 c2 = closing(result[num], previous[num], num, k)                 if c2 == 1:                     break         elif num == 3:             if abs(y[2][3][rand] - y[2][2][rand]) < err:                 result[num] = simps(y[2][3][0:max_count], u, dx=step)                 previous[num] = simps(y[2][2][0:max_count], u, dx=step)                 c3 = closing(result[num], previous[num], num, k)                 if c3 == 1:                     break         elif num == 4:             if abs(y[3][3][rand] - y[3][2][rand]) < err:                 result[num] = simps(y[3][3][0:max_count], u, dx=step)                 previous[num] = simps(y[3][2][0:max_count], u, dx=step)                 c4 = closing(result[num], previous[num], num, k)                 if c4 == 1:                     break         elif num == 5:             if abs(y[4][3][rand] - y[4][2][rand]) < err:                 result[num] = simps(y[4][3][0:max_count], u, dx=step)                 previous[num] = simps(y[4][2][0:max_count], u, dx=step)                 c5 = closing(result[num], previous[num], num, k)                 if c5 == 1:                     break         elif num == 6:             if abs(y[5][3][rand] - y[5][2][rand]) < err:                 result[num] = simps(y[5][3][0:max_count], u, dx=step)                 previous[num] = simps(y[5][2][0:max_count], u, dx=step)                 c6 = closing(result[num], previous[num], num, k)                 if c6 == 1:                     break          for i in range(max_count):  # Setting current Aitken's iteration to first iteration             if num == 1:                 x_iter[0][i] = y[0][3][i]             elif num == 2:                 x_iter[1][i] = y[1][3][i]             elif num == 3:                 x_iter[2][i] = y[2][3][i]             elif num == 4:                 x_iter[3][i] = y[3][3][i]             elif num == 5:                 x_iter[4][i] = y[4][3][i]             elif num == 6:                 x_iter[5][i] = y[5][3][i]      # If iteration limit reached, sets process closing condition to 1 and returns to main     end[num] = 1     return  if __name__ == '__main__':     i = 0      # Define and begin processes, one for each equation     p1 = mp.Process(target=f_u, args=(u, 1))     p2 = mp.Process(target=f_u, args=(u, 2))     p3 = mp.Process(target=f_u, args=(u, 3))     p4 = mp.Process(target=f_u, args=(u, 4))     p5 = mp.Process(target=f_u, args=(u, 5))     p6 = mp.Process(target=f_u, args=(u, 6))      p1.start()     p2.start()     p3.start()     p4.start()     p5.start()     p6.start()      # While any equation is running, update and show the plot every second     while end[1] != 1 or end[2] != 1 or end[3] != 1 or end[4] != 1 or end[5] != 1 or end[6] != 1:         animation.FuncAnimation(fig, animate, interval=1000, blit=True)         plt.pause(1)         plt.show(block=False)         plt.pause(1)      # Joining and closing each process when complete     p1.join()     p1.terminate()     p2.join()     p2.terminate()     p3.join()     p3.terminate()     p4.join()     p4.terminate()     p5.join()     p5.terminate()     p6.join()     p6.terminate()      end = datetime.datetime.now()   # Record time end      print("Runtime: " + str(end - start))      # Updates and plots final result     animate(i)     plt.show()      exit(0) 

Hope you can help me out. Thanks in advance.

Is there any hook or filter that user data, specifically email address, is passed through on new order creation?

I want to create a function/module that corrects common email typos on new orders.

For example to auto correct gmail.con, hotmail.con to .com, and many many more common typos.

Is there any hook or filter that user data, specifically email address, is passed through on new order creation, so that we can modify it before it’s inserted into the database?

Is it possible to cast spells on places that can be seen through a familiar? [duplicate]

This question already has an answer here:

  • How does line of sight for spells work, accounting for familiars? 1 answer

A wizard is inside a room and their familiar is outside of it. An enemy is also outside, in front of the familiar. Would the mage be able to cast, for example, frostbite

You cause numbing frost to form on one creature that you can see within range …[]…

on that target? I’m aware looking through your familiar is an action, but lets assume the wizard is hasted so it has 2 actions.

EDIT: As @Revolver_Ocelot mentioned, haste cant be used to cast an spell, so lets just assume the wizard multiclassed into a fighter and has action surge

How degrade SMB through CMD

Is there a way for downgrade the SMB protocol to SMBv1 through the CMD console?

I trying share my “program.exemple” using “impacket-smbserver” but this error is displayed:

“You can’t connect to the file share because it’s not secure. This share requires the obsolete SMB1 protocol…”

Any help is more than grateful.

Thanks.

How to make a variable-labelling function “pass through” derivatives and other functions

I have already made a function which takes any expression and replaces it with one whose variables are identical, but replaced by those with a chosen subscript (inside the “{}” part goes a set of variables which I don’t want to be labelled).

At[expr_, i_] := Module[{varsD},   varsD = Variables[expr],{}];   ReplaceAll[expr, # -> Subscript[#, i] & /@ varsD] ] 

However, what I am noticing is that, when I take functions (such as sines or derivatives etc), that the whole function is subscripted rather than just the variable.

enter image description here

I want them both to be $ \sin(\theta _l)$ and $ \dfrac{\partial F_l}{\partial y}$ . I need some way of ensuring that this only happens to the function at the highest depth, but I am not sure how to do this.

Could somebody point me in the right direction?