This is my first small python task when i found out about speedtest-cli.
import speedtest
q = 1
while q == 1:
st = speedtest.Speedtest()
option = int(input("What do you want to test:\n 1)Download Speed\n 2)Upload Speed \n 3)Ping \n Please enter the number here: "))
if option == 1:
print(st.download())
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
elif option == 2:
print(st.upload())
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
elif option == 3:
servernames =[]
st.get_servers(servernames)
print(st.results.ping)
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
else:
print("Please enter the correct choice")
else:
print("Test is ended")
i am just a beginner so i could't find any way to shorten this code. Any tips would be helpful :)
If you don't care about execution time but only about code length:
import speedtest
q = 1
while q == 1:
st = speedtest.Speedtest()
st.get_servers([])
tst_results = [st.download(), st.upload(), st.results.ping]
option = int(input("What do you want to test:\n 1)Download Speed\n 2)Upload Speed \n 3)Ping \n Please enter the number here: "))
if option >= 1 and option <= 3:
print(tst_results[option-1])
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
else:
print("Please enter the correct choice")
else:
print("Test is ended")
Did not really make it smarter, just shorter by creating a list and take option as index in the list
First, you can look at this line:
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
this line happens on each of the options, so we don't need to repeat it. instead you can add it at the end of the "while" clause. also add "continue" to your "else" clause to avoid asking this when wrong input is entered.
also, the "else" clause for the "while" loop is not needed
e.g:
import speedtest
q = 1
while q == 1:
st = speedtest.Speedtest()
option = int(input("What do you want to test:\n 1)Download Speed\n 2)Upload Speed \n 3)Ping \n Please enter the number here: "))
if option == 1:
print(st.download())
elif option == 2:
print(st.upload())
elif option == 3:
servernames =[]
st.get_servers(servernames)
print(st.results.ping)
else:
print("Please enter the correct choice")
continue
q = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))
print("Test is ended")
second, there is an error in your logic. your prompt says enter 1 to continue or 2 to quit, and indeed when you enter 2 the loop will end, but also when the user enters 3 or any other number. Even worse, if a user will enter a character that is not a number or nothing at all, they will get an exception. For this we use try-except clauses. another way to do this kind of loop is using "while "True" and then using "break" to exit.
while True:
... your code here ...
q = input("enter 1 or 2:")
try:
q = int(q)
except ValueError:
print("Invalid input")
if q == 2:
break
print("Test is ended")
This is helpful to you if:
you are a pentester
or you (for some other arbitrary reason) are looking for a way to have a very small python script
Disclaimer: I do not recommend this, just saying it may be helpful.
Steps
Replace all \n (newlines) by \\n
Replace two (or four) spaces (depending on your preferences) by \\t
Put """ (pythons long quotation marks) around it
Put that oneline string into the exec() function (not recommended, but do it if you want to)
Applied to this questions code
exec("""import speedtest\n\nq = 1\nwhile q == 1:\n\t\tst = speedtest.Speedtest()\n\t\toption = int(input("What do you want to test:\n 1)Download Speed\n 2)Upload Speed \n 3)Ping \n Please enter the number here: "))\n\t\tif\toption == 1:\n\t\t\t\tprint(st.download())\n\t\t\t\tq = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))\n\n\t\telif option == 2:\n\t\t\t\tprint(st.upload())\n\t\t\t\tq = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))\n\n\t\telif option == 3:\n\t\t\t\tservernames =[]\n\t\t\t\tst.get_servers(servernames)\n\t\t\t\tprint(st.results.ping)\n\t\t\t\tq = int(input("Enter '1' if you want to continue or Enter '2' if you want to stop the test"))\n\n\t\telse:\n\t\t\t\tprint("Please enter the correct choice")\nelse:\n\t\tprint("Test is ended")\n""")
I have found the solution to that by using the sys module. i suppose this could work if a wrong data is input.
import speedtest
import sys
#Loop for options
while True:
st = speedtest.Speedtest()
option = input("What do you want to test:\n 1)Download Speed\n 2)Upload Speed \n 3)Ping \n Please enter the number here: ")
#To check if the input value is an integer
try:
option = int(option)
except ValueError:
print("Invalid input")
continue
if option == 1:
print(st.download())
elif option == 2:
print(st.upload())
elif option == 3:
servernames =[]
st.get_servers(servernames)
print(st.results.ping)
else:
print("Please enter the correct choice: ")
continue
q = 0
#Choice to continue or to end
while (q != 1) or (q != 2):
q = input("Enter '1' if you want to continue or Enter '2' if you want to stop the test: ")
try:
q = int(q)
except ValueError:
print("Invalid input")
continue
if q == 1:
break
elif q == 2:
sys.exit("Test is ended")
else:
print("Invalid input")
continue
Related
I complete all the steps that I am given by my code and finally it asks 'again (Y or N)'. (I have written an except argument that will prevent someone from ending the code by typing a wrong answer) When they input the wrong answer it starts the user back to the top of the code. I would like it to output what step the user was already on.
CODE:
while True:
try:
choice = int(input("ONLY CHOOSE 1 AS THERE IS NO OTHER CHOICE: "))
assert choice == 1
if choice == (1):
userInp = input("TYPE: ")
words = userInp.split()
start_count = 0
for word in words:
word = word.lower()
if word.startswith("u"):
start_count += 1
print(f"You wrote {len(words)} words.")
print(f"You wrote {start_count} words that start with u.")
again = str(input("Again? (Y or N) "))
again = again.upper()
if again == "Y":
continue
elif again == "N":
break
except AssertionError:
print("Please type a given option.")
except ValueError:
print("Please type a given option.")
EDIT:
So I have made some progress but I have one last problem
CODE:
while True:
try:
choice = int(input("ONLY CHOOSE 1 AS THERE IS NO OTHER CHOICE: "))
assert choice == 1
if choice == (1):
userInp = input("TYPE: ")
words = userInp.split()
start_count = 0
for word in words:
word = word.lower()
if word.startswith("u"):
start_count += 1
print(f"You wrote {len(words)} words.")
print(f"You wrote {start_count} words that start with u.")
while True:
again = input("again? (Y or N)")
if again not in "yYnN":
continue
break
if again == "Y":
continue
elif again == "N":
break
except AssertionError:
print("Please type a given option.")
except ValueError:
print("Please type a given option.")
The problem is that the variable 'again' is not defined when outside of the while loop (outside the while loop that is in the while loop). How do I fix this?
You could create another loop (inside this main loop) where you are asking for input, and have a try block there, so that it loops just the section where the user is giving input until they give correct input.
I have seen many answers to this question but am looking for something very specific. What I need to accomplish (in pseudo code) is this:
> FOR every ITEM in DICTIONARY, DO:
> PROMPT user for input
> IF input is integer
> SET unique-variable to user input
I'm very new to Python so the code may not be proper, but here is what I have:
def enter_quantity():
for q in menu:
quantities[q] = int(input("How many orders of " + str(q) + "?: "))
So this does everything but evaluate the user input. The problem I'm having is if the input is incorrect, I need to re-prompt them for the same item in the top-level for loop. So if it's asking "How many slices of pizza?" and the user inputs "ten", I want it to say "Sorry that's not a number" and return to the prompt again of "How many slices of pizza?".
Any/all ideas are appreciated. Thanks!
My final solution:
def enter_quantity():
for q in menu:
booltest = False
while booltest == False:
inp = input("How many orders of " + str(q) + "?: ")
try:
int(inp)
booltest = True
except ValueError:
print (inp + " is not a number. Please enter a nermic quantity.")
quantities[q] = int(inp)
You need a while loop with a try/except to verify the input:
def enter_quantity():
for q in menu:
while True:
inp = input("How many orders of {} ?: ".format(q))
try:
inp = int(inp) # try cast to int
break
except ValueError:
# if we get here user entered invalid input so print message and ask again
print("{} is not a number".format(inp))
continue
# out of while so inp is good, update dict
quantities[q] = inp
This bit of code is a little more useful if a menu is added otherwise it crashes at the first hurdle. I also added a dictionary to store the input values.
menu = 'pizza', 'pasta', 'vino'
quantities = {}
def enter_quantity():
for q in menu:
while True:
if q == 'pizza':
inp = input(f"How many slices of {q} ?: ")
elif q == 'pasta':
inp = input(f"How many plates of {q} ?: ")
elif q == 'vino':
inp = input(f"How many glasses of {q} ?: ")
try:
inp = int(inp) # try cast to int
break
except ValueError:
# exception is triggered if invalid input is entered. Print message and ask again
print("{} is not a number".format(inp))
continue
# while loop is OK, update the dictionary
quantities[q] = inp
print(quantities)
Then run the code from this command:
enter_quantity()
I'm writing this program that is basically a limited calculator. I'm trying to make it so that if the user enters let's say "Power" instead of the number 1 for the desired mode, it prints out "Invalid selection". The same goes if they attempt to write "Quadratics" instead of 2 and so on for the rest.
#CALCULATOR
print("MY CALCULATOR")
print("1. Powers")
print("2. Quadratics")
print("3. Percents")
print("4. Basic Ops")
choice = int(input("Please enter your desired mode: "))
if choice == 1:
base = int(input("Enter the base: "))
exponent = int(input("Enter the exponent: "))
power = base**exponent
if choice == 2:
print("Please enter the values for A/B/C: ")
a = int(input("A: "))
b = int(input("B: "))
c = int(input("C: "))
I tried doing:
if choice not == 1:
print("Invalid Selection")
and
if choice not 1:
print("Invalid Selection")
but they don't seem to work. If you could please tell me what I am doing wrong. Thank you.
not is not a function. It is an operator.
The correct usage is to put it before an expression:
if not (choice == 1):
However in this case, it's much better to use != (not equal) instead:
if choice != 1:
Sorry quite new to python in the grand scheme of things, basically I want to be able to have an input screen that users have a selection of choices, when a choice has been made a command will run, then give the user the option to go back to the choices to choose another one, or to exit the program completely.
Currently I have
print ("1.")
print ("2.")
print ("3.")
errorch=0
while not errorch :
try :
choice = int ( input('Please Choose from Options below :') )
errorch = 1
except ValueError as e :
print ("'%s' is not a valid integer." % e.args[0].split(": ")[1])
if choice == 1:
print ("CODE:")
elif choice == 2:
print ("OTHER:")
elif choice == 3:
print ("OTHER:")
else:
print ("Invalid Choice. Please Try Again:")
k=input('Press close to exit')
In each choice I do have code that runs, but to save space I have omitted this
Use a while loop.
while True: # this loop runs forever
print("1.")
...
print("4.") # this is for exit
# get input
if choice == 1:
...
# if choice is 4, then break out of this loop
elif choice == 4:
break # breaks out of the loop
else:
...
You can just wrap the whole thing in another while loop:
while True:
...
if k.lower() == "close":
break
You can use this same form to make your existing loop neater, removing the errorch flag:
while True:
try:
choice = int(input('Please Choose from Options below :'))
except ValueError as e :
print ("'%s' is not a valid integer." % e.args[0].split(": ")[1])
else:
break
I am using python 2.6.6
I am simply trying to restart the program based on user input from the very beginning.
thanks
import random
import time
print "You may press q to quit at any time"
print "You have an amount chances"
guess = 5
while True:
chance = random.choice(['heads','tails'])
person = raw_input(" heads or tails: ")
print "*You have fliped the coin"
time.sleep(1)
if person == 'q':
print " Nooo!"
if person == 'q':
break
if person == chance:
print "correct"
elif person != chance:
print "Incorrect"
guess -=1
if guess == 0:
a = raw_input(" Play again? ")
if a == 'n':
break
if a == 'y':
continue
#Figure out how to restart program
I am confused about the continue statement.
Because if I use continue I never get the option of "play again" after the first time I enter 'y'.
Use a continue statement at the point which you want the loop to be restarted. Like you are using break for breaking from the loop, the continue statement will restart the loop.
Not based on your question, but how to use continue:
while True:
choice = raw_input('What do you want? ')
if choice == 'restart':
continue
else:
break
print 'Break!'
Also:
choice = 'restart';
while choice == 'restart':
choice = raw_input('What do you want? ')
print 'Break!'
Output :
What do you want? restart
What do you want? break
Break!
I recommend:
Factoring your code into functions; it makes it a lot more readable
Using helpful variable names
Not consuming your constants (after the first time through your code, how do you know how many guesses to start with?)
.
import random
import time
GUESSES = 5
def playGame():
remaining = GUESSES
correct = 0
while remaining>0:
hiddenValue = random.choice(('heads','tails'))
person = raw_input('Heads or Tails?').lower()
if person in ('q','quit','e','exit','bye'):
print('Quitter!')
break
elif hiddenValue=='heads' and person in ('h','head','heads'):
print('Correct!')
correct += 1
elif hiddenValue=='tails' and person in ('t','tail','tails'):
print('Correct!')
correct += 1
else:
print('Nope, sorry...')
remaining -= 1
print('You got {0} correct (out of {1})\n'.format(correct, correct+GUESSES-remaining))
def main():
print("You may press q to quit at any time")
print("You have {0} chances".format(GUESSES))
while True:
playGame()
again = raw_input('Play again? (Y/n)').lower()
if again in ('n','no','q','quit','e','exit','bye'):
break
You need to use random.seed to initialize the random number generator. If you call it with the same value each time, the values from random.choice will repeat themselves.
After you enter 'y', guess == 0 will never be True.