python中来自用户matplotlib的输入函数

Question

我正在尝试制作一个可以绘制用户编写函数的程序。 该图来自参数函数。 我在python中使用matplotlib，但我可以做到这一点，即用户将convert转换为matplotlib使用的函数。 你能帮我吗？ 还是可以建议我其他python的库？ 这是我的代码以及我尝试过的内容：

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()

在第一个示例中，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matplotlib时，出现了一个错误，并且是因为用户输入是一个字符串，而不是所需变量的种类。

谢谢您的帮助，对不起我的英语。

Answer 1

我修改了这个pyparsing程序 ，以接受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，此程序执行以下操作：

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.        ]

因此，基本上， compute功能将评估简单表达式，并在使用t提供linspace（var）。 如果需要，可以扩展它。 我认为您可以从这里合并所需的内容以使程序正常工作。 :)