如何将 lambda 函数存储在 Python 的字典中？

Question

Someone shared their code and I saw a bunch of functions that were stored in what seemed to me to be a dictionary and. So, I liked the idea and I borrowed them. The code that the person wrote it in was in JS, and I work with Python, so I translated the code into Python.

Here is what that person wrote in JS:

EasingFunctions = {
  // no easing, no acceleration
  linear: t => t,
  // accelerating from zero velocity
  easeInQuad: t => t * t,
  // decelerating to zero velocity
  easeOutQuad: t => t * (2 - t),
  // acceleration until halfway, then deceleration
  easeInOutQuad: t => t < .5 ? 2 * t * t : -1 + (4 - 2 * t) * t,
  // accelerating from zero velocity 
  easeInCubic: t => t * t * t,
  // decelerating to zero velocity 
  easeOutCubic: t => (--t) * t * t + 1,
  // acceleration until halfway, then deceleration 
  easeInOutCubic: t => t < .5 ? 4 * t * t * t : (t - 1) * (2 * t - 2) * (2 * t - 2) + 1,
  // accelerating from zero velocity 
  easeInQuart: t => t * t * t * t,
  // decelerating to zero velocity 
  easeOutQuart: t => 1 - (--t) * t * t * t,
  // acceleration until halfway, then deceleration
  easeInOutQuart: t => t < .5 ? 8 * t * t * t * t : 1 - 8 * (--t) * t * t * t,
  // accelerating from zero velocity
  easeInQuint: t => t * t * t * t * t,
  // decelerating to zero velocity
  easeOutQuint: t => 1 + (--t) * t * t * t * t,
  // acceleration until halfway, then deceleration 
  easeInOutQuint: t => t < .5 ? 16 * t * t * t * t * t : 1 + 16 * (--t) * t * t * t * t
}

And it works fine if you ran this code. However in the code I translated, it gave me an error saying that I miss a paranthesis, comma, or a colon. Here is the code:

EasingFunctions = {
  # no easing, no acceleration
  linear: lambda t : t,
  # accelerating from zero velocity
  easeInQuad: lambda t : t ** 2,
  # decelerating to zero velocity
  easeOutQuad: lambda t : t * (2-t),
  # acceleration until halfway, then deceleration
  easeInOutQuad: (lambda t : t = (2*(t**2)) if t < 0.5 else ((-1+(4-2*t)) * t)),
  # accelerating from zero velocity 
  easeInCubic: lambda t : t * t * t,
  # decelerating to zero velocity 
  easeOutCubic: lambda t : (t-1) * t * t + 1, 
  # acceleration until halfway, then deceleration 
  easeInOutCubic: lambda t : t = 4*t*t*t if t < 0.5 else (t - 1) * (2 * t - 2) * (2 * t - 2) + 1,
  # accelerating from zero velocity 
  easeInQuart: lambda t : t ** 4,
  # decelerating to zero velocity 
  easeOutQuart: lambda t : 1 - (t-1) * t * t * t,
  # acceleration until halfway, then deceleration
  easeInOutQuart: lambda t : t = 8 * t * t * t * t if t < 0.5 else 1 - 8 * (t) * t * t * t
  # accelerating from zero velocity
  easeInQuint: lambda t : t ** 5,
  # decelerating to zero velocity
  easeOutQuint: lambda t : 1 + (t-1) * t * t * t * t,
  # acceleration until halfway, then deceleration 
  easeInOutQuint: lambda t : t = 16 * t * t * t * t * t if t < 0.5 else 1 + 16 * (t-1) * t * t * t * t  
}

And what confused me is that the error was indicated to be the first key value that had an if statement in it. I thought that this was allowed in Python, what is wrong with the code?

Answer 1

As you mentioned in the comments that you still can't figure out how to do it using string keys for dictionary, I'm posting this answer. Though, it was partially mentioned in the comments how to do this.

a = {
    'linear': lambda t: t,
    'easeInQuad': lambda t: t ** 2,
    'easeOutQuad': lambda t: t * (2-t),
    'easeOutQuint': lambda t: 1 + (t - 1) * t * t * t * t,
}

print(a['linear'](69))
print(a['easeInQuad'](69))
print(a['easeOutQuad'](69))
print(a['easeOutQuint'](69))

Result:

69
4761
-4623
1541364229

Again, as mentioned in comments, Python doesn't support -- operation. Hope this helps.