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Check if the input is integer float or string in python

I am a beginner in python and I am currently working on a calculator not like this:
Enter Something: add
"Enter 1 number : 1"
"Enter 2 number : 3"
The answer is 5
not like that or using eval()
I Want to create a calculator where they input something like this: "add 1 3" and output should be 4.
but I have to check that the first word is a string 2nd is a integer or float and 3rd is also number
I have created a script but I have one problem that I don't know how to check if the input is a integer or string or float I have used isdigit() it works but it doesn't count negative numbers and float as a number I have also used isinstance() but it doesn't work and thinks that the input is a integer even when its a string and I don't know how to use the try and except method on this script
while True:
exitcond = ["exit","close","quit"]
operators =["add","subtract","multiply","divide"]
uinput = str(input())
lowereduin = uinput.lower()
splited = lowereduin.split(" ")
if lowereduin in exitcond:
break
if splited[0] == operators[0]:
if isinstance(splited[1],int) == True:
if isinstance(splited[2] , int) == True:
result = int(splited[1]) + int(splited[2])
print(result)
else:
print("enter a number")
else:
print("enter a number")
and when I run this script and type add 1 3 its says enter a number and when I only type add its give this error
Traceback (most recent call last):
File "C:\Users\Tyagiji\Documents\Python Projects\TRyinrg differet\experiments.py", line 11, in <module>
if isinstance(splited[1],int) == True:
IndexError: list index out of range
Can someone tell me what's this error and if this doesn't work can you tell me how to use try: method on this script.

You can try the following approach and play with type checking
import operator
while True:
exitcond = ["exit","close","quit"]
operators ={"add": operator.add,"subtract":operator.sub,"multiply": operator.mul,"divide":operator.truediv}
uinput = str(input())
lowereduin = uinput.lower()
splited = lowereduin.split(" ")
if lowereduin in exitcond or (len(splited) !=3):
break
try:
if splited[0] not in operators.keys():
raise ValueError(f"{splited[0]} not in {list(operators.keys())}")
op = operators.get(splited[0])
val = op(
*map(int, splited[1:])
)
print(val)
except (ValueError, ZeroDivisionError) as err:
print(err)
break

Building on Deepak's answer. A dictionary of operator names to functions is a good approach. And you can add to splits until you have enough numbers to proceed.
import operator as op
while True:
exitcond = ["exit","close","quit"]
operators = {"add": op.add,"subtract": op.sub, "multiply": op.mul, "divide": op.truediv}
splits = str(input()).lower().split()
if any(part in exitcond for part in splits):
break
while len(splits) < 3:
splits.append(input('Enter number: '))
try:
print(operators[splits[0]](*map(lambda x: float(x.replace(',','')), splits[1:3])))
except ZeroDivisionError:
print("Can't divide by 0")
except:
print('Expected input: add|subtract|multiply|divide [number1] [number2] -- or -- exit|quit|close')
Things to note, the lambda function removes all commas , as they fail for floats, and then converts the all number strings to floats. So, the answer will always be a float, which opens the can of worms that add 1.1 2.2 won't be exactly 3.3 due to the well documented issues with floating point arithmatic and computers

Issue in providing special character as input in python [duplicate]

I want to get a string from a user, and then to manipulate it.
testVar = input("Ask user for something.")
Is there a way for testVar to be a string without me having the user type his response in quotes? i.e. "Hello" vs. Hello
If the user types in Hello, I get the following error:
NameError: name 'Hello' is not defined

Use raw_input() instead of input():
testVar = raw_input("Ask user for something.")
input() actually evaluates the input as Python code. I suggest to never use it. raw_input() returns the verbatim string entered by the user.

The function input will also evaluate the data it just read as python code, which is not really what you want.
The generic approach would be to treat the user input (from sys.stdin) like any other file. Try
import sys
sys.stdin.readline()
If you want to keep it short, you can use raw_input which is the same as input but omits the evaluation.

We can use the raw_input() function in Python 2 and the input() function in Python 3.
By default the input function takes an input in string format. For other data type you have to cast the user input.
In Python 2 we use the raw_input() function. It waits for the user to type some input and press return and we need to store the value in a variable by casting as our desire data type. Be careful when using type casting
x = raw_input("Enter a number: ") #String input
x = int(raw_input("Enter a number: ")) #integer input
x = float(raw_input("Enter a float number: ")) #float input
x = eval(raw_input("Enter a float number: ")) #eval input
In Python 3 we use the input() function which returns a user input value.
x = input("Enter a number: ") #String input
If you enter a string, int, float, eval it will take as string input
x = int(input("Enter a number: ")) #integer input
If you enter a string for int cast ValueError: invalid literal for int() with base 10:
x = float(input("Enter a float number: ")) #float input
If you enter a string for float cast ValueError: could not convert string to float
x = eval(input("Enter a float number: ")) #eval input
If you enter a string for eval cast NameError: name ' ' is not defined
Those error also applicable for Python 2.

If you want to use input instead of raw_input in python 2.x,then this trick will come handy
if hasattr(__builtins__, 'raw_input'):
input=raw_input
After which,
testVar = input("Ask user for something.")
will work just fine.

testVar = raw_input("Ask user for something.")

My Working code with fixes:
import random
import math
print "Welcome to Sam's Math Test"
num1= random.randint(1, 10)
num2= random.randint(1, 10)
num3= random.randint(1, 10)
list=[num1, num2, num3]
maxNum= max(list)
minNum= min(list)
sqrtOne= math.sqrt(num1)
correct= False
while(correct == False):
guess1= input("Which number is the highest? "+ str(list) + ": ")
if maxNum == guess1:
print("Correct!")
correct = True
else:
print("Incorrect, try again")
correct= False
while(correct == False):
guess2= input("Which number is the lowest? " + str(list) +": ")
if minNum == guess2:
print("Correct!")
correct = True
else:
print("Incorrect, try again")
correct= False
while(correct == False):
guess3= raw_input("Is the square root of " + str(num1) + " greater than or equal to 2? (y/n): ")
if sqrtOne >= 2.0 and str(guess3) == "y":
print("Correct!")
correct = True
elif sqrtOne < 2.0 and str(guess3) == "n":
print("Correct!")
correct = True
else:
print("Incorrect, try again")
print("Thanks for playing!")

This is my work around to fail safe in case if i will need to move to python 3 in future.
def _input(msg):
return raw_input(msg)

The issue seems to be resolved in Python version 3.4.2.
testVar = input("Ask user for something.")
Will work fine.

Python does not accept my input as a real number

# Math Quizzes
import random
import math
import operator
def questions():
# Gets the name of the user
name= ("Alz")## input("What is your name")
for i in range(10):
#Generates the questions
number1 = random.randint(0,100)
number2 = random.randint(1,10)
#Creates a Dictionary containg the Opernads
Operands ={'+':operator.add,
'-':operator.sub,
'*':operator.mul,
'/':operator.truediv}
#Creast a list containing a dictionary with the Operands
Ops= random.choice(list(Operands.keys()))
# Makes the Answer variable avialabe to the whole program
global answer
# Gets the answer
answer= Operands.get(Ops)(number1,number2)
# Makes the Sum variable avialbe to the whole program
global Sum
# Ask the user the question
Sum = ('What is {} {} {} {}?'.format(number1,Ops,number2,name))
print (Sum)
global UserAnswer
UserAnswer= input()
if UserAnswer == input():
UserAnswer= float(input())
elif UserAnswer != float() :
print("Please enter a correct input")
def score(Sum,answer):
score = 0
for i in range(10):
correct= answer
if UserAnswer == correct:
score +=1
print("You got it right")
else:
return("You got it wrong")
print ("You got",score,"out of 10")
questions()
score(Sum,answer)
When I enter a float number into the console the console prints out this:
What is 95 * 10 Alz?
950
Please enter a correct input
I'm just curious on how I would make the console not print out the message and the proper number.

this is a way to make sure you get something that can be interpreted as a float from the user:
while True:
try:
user_input = float(input('number? '))
break
except ValueError:
print('that was not a float; try again...')
print(user_input)
the idea is to try to cast the string entered by the user to a float and ask again as long as that fails. if it checks out, break from the (infinite) loop.

You could structure the conditional if statement such that it cause number types more than just float
if UserAnswer == input():
UserAnswer= float(input())
elif UserAnswer != float() :
print("Please enter a correct input")

Trace through your code to understand why it doesn't work:
UserAnswer= input()
This line offers no prompt to the user. Then it will read characters from standard input until it reaches the end of a line. The characters read are assigned to the variable UserAnswer (as type str).
if UserAnswer == input():
Again offer no prompt to the user before reading input. The new input is compared to the value in UserAnswer (which was just entered on the previous line). If this new input is equal to the previous input then execute the next block.
UserAnswer= float(input())
For a third time in a row, read input without presenting a prompt. Try to parse this third input as a floating point number. An exception will be raised if this new input can not be parsed. If it is parsed it is assigned to UserAnswer.
elif UserAnswer != float() :
This expression is evaluated only when the second input does not equal the first. If this is confusing, then that is because the code is equally confusing (and probably not what you want). The first input (which is a string) is compared to a newly created float object with the default value returned by the float() function.
Since a string is never equal to a float this not-equals test will always be true.
print("Please enter a correct input")
and thus this message is printed.
Change this entire section of code to something like this (but this is only a representative example, you may, in fact, want some different behavior):
while True:
try:
raw_UserAnswer = input("Please enter an answer:")
UserAnswer = float(raw_UserAnswer)
break
except ValueError:
print("Please enter a correct input")

python quiz validation not working

The validation doesnt work. im not sure why, is there a way to validate a string. The questions asked are endless i need 10 questions to be asked
import random
name=(input("Please enter your name"))
print("welcome",name,"the arithmetic is about to start")
question=0
while question<10:
number=random.randint(1,10)
numbers=random.randint(1,10)
arith=random.choice("+" "-" "/")
if arith=="+":
print(number,arith,numbers)
answer=number+numbers
if arith=="-":
print(number,arith,numbers)
answer=number-numbers
if arith=="/":
print(number,arith,numbers)
answer=number/numbers
while True:
try:
usersanswer= int(input())
except ValueError:
print ("That is not a valid answer")
continue
if usersanswer==answer:
print("correct")
break
else:
print("incorrct")
The validation doesnt work. im not sure why, is there a way to validate a string

I've taking silentphoenix's answer and made it somewhat more pythonic and six'ed.
You should almost never use python2's input, because on top of being massive security hole, it sometimes does things that can be...rather unexpected.
import random
import operator # contains the python operators as functions
try:
input = raw_input # rebind raw_input to input, if it exists
# so I can just use input :P
except NameError:
pass
name = input("Hi, what is your name?\n")
print("Hi {} let's get started! Question 1".format(name))
#Get out of the habit of using string concatenation and use string
#formatting whenever possible. Strings are *immutable*;
#concatenation has to produce a lot temporary strings and is *slow*
#str.join and str.format are almost always better ideas.
#Python does not have a switch-case, so emulating one with a dictionary
operator_mapping = {'+': operator.add,
'-': operator.sub,
'*': operator.mul,
#'/': operator.truediv, #hey, division exists.
#But if you want division to actually work, you'll
#have to introduce a fudge factor :P
}
for i in range(10): # If you're just going for 10 iterations, it should be a for loop
# Brevity :P This is a list comprehension
first_number, second_number = [random.randint(1,10) for _ in range(2)]
oper = random.choice(list(operator_mapping))
answer = operator_mapping[oper](first_number, second_number)
while int(input("{} {} {} = ".format(first_number, oper, second_number))) != answer:
#while abs(float(input("{} {} {} = ".format(first_number, oper, second_number)))-answer) < 0.001: if you want truediv.
print('Wrong answer! try again!')
#If I've left the loop, user has given correct (enough) answer
if i <9: # all but last
print('Well done! Now onto question number {0}'.format(i+2))
print('Well done! You are done!')

In the third line, you ask for input. But a name is a string, so you need raw_input. raw_input takes strings, input only takes numerical values.
Python 2.7 getting user input and manipulating as string without quotations
Nowhere in your code do you update the variable questions, which I am guessing is a counter. You have to update that whenever a question is asked, using question += 1.
Finally, your code at the end does not really make sense. Based off the code, it checks for whether or not it is a string, but then compares it to the answer regardless. The if statement needs to be within the try.
The else statement does not match any outer indentation.
Finally, because of the while True: your code will never exit the loop unless the answer is wrong. At the point the entire program terminates. I see what kind of program you are trying to write, but the parameters for random number generation have to be within some kind of a while question <= 10 loop. As of now, only two lines in the program are being affected by that first while loop.
EDIT: I am working on a good example code. Hopefully this answer will help until I can finish it.
EDIT: Here is code that shows how it works within a while loop.
import random
from random import randint
name = raw_input("Hi, what is your name?\n") # Asks for name
print "Hi " +name+ " let's get started!"
score_count = 0
question_count = 0 # creates counter
while question_count <= 10: # Everything MUST BE WITHIN THIS LOOP
# makes numbers and operator
first_number = randint(1,10)
second_number = randint(1,10)
oper = random.choice("+""-""*")
# determines the problem
if oper == "+":
answer = first_number + second_number
print first_number,second_number,oper
elif oper == "-":
answer = first_number - second_number
print first_number,second_number,oper
elif oper == "*":
answer = first_number*second_number
print first_number, second_number, oper
user_answer = int(raw_input("Your answer: "))
if user_answer != answer:
print 'Wrong answer! try again!'
user_answer = int(raw_input('Your answer: '))
if user_answer == answer: # exits the while loop when the correct answer is given
if question_count < 10:
print 'Well done! Now onto question number {0}'.format(question_count+1)
score_count += 1
elif question_count == 10:
print 'Well done! You are done!'
score_count += 1
else:
print 'Something is wrong.'
question_count += 1 # updates the variable
# GOES BACK TO THE BEGINNING UNTIL question_count IS GREATER THAN OR EQUAL TO 10
print "Your score was: {}".format(score_count)
Happy coding! and best of luck!

hi im Nathan and I saw this post I am 5 years to late but I figured if someone on here is knew to python I have a much easier (in my opinion) way to do this in python 3, the code is below:
import random #random module automatically downloaded when you install python
name = input("Please enter your name ")
print("welcome",name,"the arithmetic is about to start")
question=0
while question<10:
number=random.randint(1,10) #creating a random number
numbers=random.randint(1,10) #creating a random number
list = ["+","-","/"] #creating a list (or sometimes called array)
arith=random.choice(list) #getting random operators from list (+,-,/)
question += 1 #basically means add one to question variable each time in loop
if arith=="+":
print(number,arith,numbers)
answer=number+numbers
elif arith=="-":
print(number,arith,numbers)
answer=number-numbers
elif arith=="/":
print(number,arith,numbers)
answer=number/numbers
answer = int(answer)
#from HERE
useranswer = "initialising this variable"
while useranswer == "initialising this variable":
try:
usersanswer= int(input())
if usersanswer==answer:
print("correct")
break
else:
print("incorrect")
except ValueError:
print ("That is not a valid answer")
#to HERE it is input validation this takes a while to explain in just commenting
#but if you dont know what this is then copy this link https://youtu.be/EG69-5U2AfU
#and paste into google for a detailed video !!!!!!
I hope this helps and is a more simplified commented bit of code to help you on your journey to code in python

Else is being displayed even if correct

I've been having a problem in my multiplication game where else would be displayed even if the answer was correct.
Here is a sample of the code:
for num in range(0,1):
number1 = random.randint(2,5)
number2 = random.randint(2,5)
answer = number1*number2
guess = input("What is %d x %d? " % (number1, number2))
if guess == answer:
print('Correct')
else:
print('Incorrect')

In Python 3.x, Input returns a str, vs. in Python 2.x where it attempted to evaluate the input as python code. And str == int always returns False, and doesn't throw an exception. You'd need to change your code to:
if guess == str(answer):
if you'd like to avoid throwing exceptions if the input isn't actually a number, or
gess = int(input(...))
if you intend to actually use guess as a number later on, but will then have to handle what happens if the user enters not a number.