python中用户matplotlib的输入函数

Question

我正在尝试制作一个可以绘制用户编写的函数的程序。该图来自参数函数。 我在 python 中使用 matplotlib，但我可以将用户写入转换为 matplotlib 使用的函数。 你能帮我吗？或者你能推荐我其他python的库吗？ 这是我的代码和我尝试过的：

import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib.widgets import CheckButtons
from numpy import *

go=False
inputX=raw_input("Write the function for X: ")
inputY=raw_input("Write the function for Y: ")
inputZ=raw_input("Write the function for Z: ")

if len(inputX)>0 and len(inputY)>0 and len(inputZ)>0:
  go=True

if go==True:
  mpl.rcParams['legend.fontsize'] = 10
  fig = plt.figure()
  ax = fig.gca(projection='3d')

  t = linspace(-4 * pi, 4 * pi, 200)

  #I would like that the user input convert in the function for each "axis"
  z = t
  x = sin(t)
  y = cos(t)

  ax.plot(x, y, z, label='Test')

  ax.legend()
  plt.show()

首先，我只是检查了每个输入的 len，但是当我尝试将输入传递给函数时，matplotlib 给了我一个错误，因为用户输入是一个字符串，而不是所需的变量类型.

感谢您的帮助，对我的英语感到抱歉。

Answer 1

我修改了this pyparsing program，接受t作为变量：

from pyparsing import Literal,CaselessLiteral,Word,Combine,Group,Optional,\
    ZeroOrMore,Forward,nums,alphas
from numpy import *
import operator

exprStack = []

def pushFirst( strg, loc, toks ):
    exprStack.append( toks[0] )
def pushUMinus( strg, loc, toks ):
    if toks and toks[0]=='-': 
        exprStack.append( 'unary -' )
        #~ exprStack.append( '-1' )
        #~ exprStack.append( '*' )

bnf = None
def BNF():
    """
    expop   :: '^'
    multop  :: '*' | '/'
    addop   :: '+' | '-'
    integer :: ['+' | '-'] '0'..'9'+
    atom    :: PI | E | real | T | fn '(' expr ')' | '(' expr ')'
    factor  :: atom [ expop factor ]*
    term    :: factor [ multop factor ]*
    expr    :: term [ addop term ]*
    """
    global bnf
    if not bnf:
        point = Literal( "." )
        e     = CaselessLiteral( "E" )
        fnumber = Combine( Word( "+-"+nums, nums ) + 
                           Optional( point + Optional( Word( nums ) ) ) +
                           Optional( e + Word( "+-"+nums, nums ) ) )
        ident = Word(alphas, alphas+nums+"_$")

        plus  = Literal( "+" )
        minus = Literal( "-" )
        mult  = Literal( "*" )
        div   = Literal( "/" )
        lpar  = Literal( "(" ).suppress()
        rpar  = Literal( ")" ).suppress()
        addop  = plus | minus
        multop = mult | div
        expop = Literal( "^" )
        pi    = CaselessLiteral( "PI" )
        t     = CaselessLiteral( "T" )

        expr = Forward()
        atom = (Optional("-") + ( pi | e | t | fnumber | ident + lpar + expr + rpar ).setParseAction( pushFirst ) | ( lpar + expr.suppress() + rpar )).setParseAction(pushUMinus) 

        # by defining exponentiation as "atom [ ^ factor ]..." instead of "atom [ ^ atom ]...", we get right-to-left exponents, instead of left-to-righ
        # that is, 2^3^2 = 2^(3^2), not (2^3)^2.
        factor = Forward()
        factor << atom + ZeroOrMore( ( expop + factor ).setParseAction( pushFirst ) )

        term = factor + ZeroOrMore( ( multop + factor ).setParseAction( pushFirst ) )
        expr << term + ZeroOrMore( ( addop + term ).setParseAction( pushFirst ) )
        bnf = expr
    return bnf

# map operator symbols to corresponding arithmetic operations
epsilon = 1e-12
opn = { "+" : operator.add,
        "-" : operator.sub,
        "*" : operator.mul,
        "/" : operator.truediv,
        "^" : operator.pow }
fn  = { "sin" : sin,
        "cos" : cos,
        "tan" : tan,
        "abs" : abs,
        "trunc" : lambda a: int(a),
        "round" : round,
        "sgn" : lambda a: abs(a)>epsilon and cmp(a,0) or 0}

def evaluateStack( s ):
    op = s.pop()
    if op == 'unary -':
        return -evaluateStack( s )
    if op in "+-*/^":
        op2 = evaluateStack( s )
        op1 = evaluateStack( s )
        return opn[op]( op1, op2 )
    elif op == "PI":
        return math.pi # 3.1415926535
    elif op == "E":
        return math.e  # 2.718281828
    elif op == "T":
        return linspace(-4 * pi, 4 * pi, 200)
    elif op in fn:
        return fn[op]( evaluateStack( s ) )
    elif op[0].isalpha():
        return 0
    else:
        return float( op )

def compute(s):   
    global exprStack
    exprStack = []
    results = BNF().parseString( s )
    val = evaluateStack( exprStack[:] )
    return val

fun=raw_input("Please input a function to be evaluated: ")
print compute(fun)

例如，该程序执行以下操作：

Please input a function to be evaluated: 2+sin(t)
[ 2.          2.12595971  2.24991296  2.36988529  2.4839656   2.59033669
  2.68730414  2.77332333  2.84702403  2.90723225  2.95298891  2.98356514
  2.99847388  2.99747765  2.98059231  2.94808684  2.90047903  2.83852724
  2.7632183   2.67575185  2.57752115  2.47009096  2.35517255  2.2345965
  2.1102835   1.98421376  1.85839548  1.73483286  1.61549416  1.50228037
  1.39699489  1.30131485  1.21676437  1.14469026  1.08624063  1.04234653
  1.01370716  1.00077873  1.00376718  1.0226249   1.05705151  1.1064986
  1.17017854  1.24707694  1.33596886  1.43543832  1.54390084  1.6596287
  1.78077842  1.90542019  2.03156855  2.15721404  2.28035523  2.39903056
  2.51134962  2.61552324  2.70989202  2.79295273  2.86338228  2.92005876
  2.96207936  2.98877473  2.99971963  2.99473972  2.97391431  2.93757514
  2.88630108  2.82090888  2.74244018  2.65214495  2.55146151  2.44199369
  2.32548523  2.20379202  2.07885255  1.95265701  1.82721561  1.70452655
  1.58654417  1.47514784  1.37211203  1.27907802  1.19752779  1.12876036
  1.07387115  1.03373451  1.00898979  1.00003115  1.0070013   1.0297892
  1.06803187  1.12112012  1.1882083   1.26822773  1.35990378  1.46177609
  1.57222193  1.68948197  1.81168833  1.93689437  2.06310563  2.18831167
  2.31051803  2.42777807  2.53822391  2.64009622  2.73177227  2.8117917
  2.87887988  2.93196813  2.9702108   2.9929987   2.99996885  2.99101021
  2.96626549  2.92612885  2.87123964  2.80247221  2.72092198  2.62788797
  2.52485216  2.41345583  2.29547345  2.17278439  2.04734299  1.92114745
  1.79620798  1.67451477  1.55800631  1.44853849  1.34785505  1.25755982
  1.17909112  1.11369892  1.06242486  1.02608569  1.00526028  1.00028037
  1.01122527  1.03792064  1.07994124  1.13661772  1.20704727  1.29010798
  1.38447676  1.48865038  1.60096944  1.71964477  1.84278596  1.96843145
  2.09457981  2.21922158  2.3403713   2.45609916  2.56456168  2.66403114
  2.75292306  2.82982146  2.8935014   2.94294849  2.9773751   2.99623282
  2.99922127  2.98629284  2.95765347  2.91375937  2.85530974  2.78323563
  2.69868515  2.60300511  2.49771963  2.38450584  2.26516714  2.14160452
  2.01578624  1.8897165   1.7654035   1.64482745  1.52990904  1.42247885
  1.32424815  1.2367817   1.16147276  1.09952097  1.05191316  1.01940769
  1.00252235  1.00152612  1.01643486  1.04701109  1.09276775  1.15297597
  1.22667667  1.31269586  1.40966331  1.5160344   1.63011471  1.75008704
  1.87404029  2.        ]

所以基本上compute 函数将评估简单的表达式，并在您使用t 时提供一个linspace (var)。如果你愿意，你可以扩展它。我认为你可以从这里合并你需要的东西来使你的程序工作。 :)