Lambda function asks for input twice,

Question

I am working on a rather complicated math project using python but keep getting stuck with one problem:

a = lambda x: input('Function in terms of x')

This works until I have to run a command like:

z = a(n)

Every time this is done it asks for an input again so I get a console that looks like:

Function in terms of x:
Function in terms of x:

I know that I should theoretically be able to get around with the following:

func = input('Function: ')
a = lambda x: func

This creates another problem: x isn't defined in the outer scope, so I added sympy symbols like so:

x = sym.Symbol('x')
func = input('Function: ')
a = lambda x: func

but then running a command like this results in something weird:

Function: 5*x +1
>>>a(10)
5*x + 1

I think this is because lambda doesn't work with sympy, but I can't think of another way to get around the problem... Thank You for any help.

The full code is as follows; the function asks for the first input at line 18, and then at line 50 where it isn't supposed too. I believe that this has to do with the fact that I use the lambda function twice.

import matplotlib.pyplot as plt
import os
import time
from mpl_toolkits.mplot3d import axes3d
from sympy import *
import numpy as np
import tkinter as tk
from colorama import init, Fore, Back, Style
import mpmath


def main():
    """
    Handling for Range and function
    """
    rng = raw_input('Minimum, Maximum: ').split(',')
    rng = [float(rng[i]) for i in range(2)]
    a = lambda x: input('Function of x: ')  # function a is the main polynomial#
    """
        2 Dimensional Graph
    """
    two_d_x = np.arange(rng[0], rng[1], abs(rng[1] - rng[0]) / 100)
    two_d_y = a(two_d_x)
    fig1 = plt.figure()
    ax1 = fig1.add_subplot(221)
    print [np.amin(two_d_x), np.amax(two_d_x), np.amin(two_d_y), np.amax(two_d_y)]
    ax1.axis([np.amin(two_d_x), np.amax(two_d_x), np.amin(two_d_y), np.amax(two_d_y)])
    ax1.plot(two_d_x, two_d_y, 'r-')
    ax1.set_title(r'$\mathit{f(x)}\in \mathbb{R}^2$')
    ax1.set_xlabel(r'$\mathit{x}$')
    ax1.set_ylabel(r'$\mathit{y}$')
    ax1.grid()
    ax1.spines['left'].set_position('zero')
    ax1.spines['right'].set_color('none')
    ax1.spines['bottom'].set_position('zero')
    ax1.spines['top'].set_color('none')
    ax1.spines['left'].set_smart_bounds(True)
    ax1.spines['bottom'].set_smart_bounds(True)
    plt.gca().set_aspect('equal', adjustable='box')
    ax1.xaxis.set_ticks_position('bottom')
    ax1.yaxis.set_ticks_position('left')
    """
        Quiver Plot of Function
    """
    ax2 = fig1.add_subplot(222)
    u, v = np.meshgrid(np.arange(rng[0], rng[1], 1),
                       np.arange(rng[0], rng[1], 1))
    ### u+vj -> w+rjf
    print False
    output = a(u + (v * 1j))
    print False
    w = output.real
    r = output.imag
    ax2.axis([np.amin(w) * 1.1, np.amax(w) * 1.1, np.amin(r) * 1.1, np.amax(r) * 1.1])
    distance = np.sqrt(((w - u) ** 2) + ((r - v) ** 2))
    quiver_plot = ax2.quiver(u, v, w, r, distance, angles='xy', scale_units='xy', scale=1, cmap=plt.cm.jet)
    plt.colorbar(quiver_plot, cmap=plt.cm.jet)
    ax2.set_title(r'$\mathit{f(x)}\in \mathbb{C}^2$')
    ax2.set_xlabel(r'$\mathit{rl}$')
    ax2.set_ylabel(r'$\mathit{im}$')
    ax2.grid()
    ax2.spines['left'].set_position('zero')
    ax2.spines['right'].set_color('none')
    ax2.spines['bottom'].set_position('zero')
    ax2.spines['top'].set_color('none')
    ax2.spines['left'].set_smart_bounds(True)
    ax2.spines['bottom'].set_smart_bounds(True)
    plt.gca().set_aspect('equal', adjustable='box')
    ax2.xaxis.set_ticks_position('bottom')
    ax2.yaxis.set_ticks_position('left')
    plt.show()


main_program_loop = True
while main_program_loop == True:
    print '| Quandri 1.0 | by: Boolean Designs\n'
    main()
    stay_loop_tp = True
    while stay_loop_tp != False:
        stay_loop_tp = raw_input("Would you like to continue using this program <yes/no>? ")
        if stay_loop_tp == 'yes' or stay_loop_tp == 'y':
            os.system('cls')
            stay_loop_tp = False
        elif stay_loop_tp == 'no' or stay_loop_tp == 'n':
            print 'Exiting Quandri...'
            time.sleep(1)
            exit()
            stay_loop_tp = False
        else:
            print "Improper Input."
            time.sleep(2)
            os.system('cls')

At which point do you actually evaluate the entered function? I don't see that happening. You call 'a(10)', which calls plain 'func', which in turn calls the input statement, and the return value of the input statement is then passed back through until the initial call. I.e., it's no surprise, that you only get back your '5*x+1'. — Thomas Kühn
– Thomas Kühn, Commented May 14, 2017 at 17:46
I cannot reproduce your repeated statement on python 2.7, it just works. — Yasin Yousif
– Yasin Yousif, Commented May 14, 2017 at 17:52

Community · Accepted Answer · 2017-05-23 10:31:10Z

1

The sympy lib has support for parsing and evaluating expression:

import sympy
from sympy.parsing.sympy_parser import parse_expr

x = sympy.Symbol('x')
expression_string = input("Function: ")
expr = parse_expr(expression_string)
expr.evalf(subs={x:10})

see this http://docs.sympy.org/dev/modules/parsing.html and this How to calculate expression using sympy in python

Edit: Thomas Kühn's answer is good, but in python2.7 raw_input must be used

f = raw_input("Function: ")
a = lambda x:eval(f) 
print(a(10))

edited May 23, 2017 at 10:31

CommunityBot

11 silver badge

answered May 14, 2017 at 17:49

Ilario Pierbattista

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6 Comments

BooleanDesigns Over a year ago

I used to use this method, but I have to parse/evaluate a meshgrid, which is longer than 32 creating another error from Sympy.

Ilario Pierbattista Over a year ago

Can you provide further details about that error? Maybe it's easier to find a work around for that instead of writing a new parser/evaluator.

BooleanDesigns Over a year ago

'code'Traceback (most recent call last): z = function.evalf(subs={x:X,y;Y}) ValueError: sequence too large; cannot be greater than 32

Ilario Pierbattista Over a year ago

What are X and Y? Scalar or vectors? That error seems to be raised by numpy: it seems that you're trying to allocate a multidimensional array with the number of dimensions greater than 32

BooleanDesigns Over a year ago

X and Y are a meshgrid for 3d graphing. So I guess that makes them scalar.

|

Thomas Kühn · Accepted Answer · 2017-05-14 17:54:07Z

0

I don't know, if it is advisable to do something like this, but here is some code that works for me on python 3.5:

func = input('Function: ')
a = lambda x: eval(func)
print(a(10))

calling this script with python, I get:

Function: x+1
11

answered May 14, 2017 at 17:54

Thomas Kühn

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1 Comment

BooleanDesigns Over a year ago

Not supported by Python 2.7. Thanks for the help though.

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