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Extract numbers from a string in python without the use of isdigit or re. tools

Let's say I have a string of integers generated by user input, where each integer is separated by a space (Code below for example)...
How can I search through that string and store each integer separately for use later on in the program? (I.E. Assigning each integer to its own variable) I can't use isdigit and cant use re tools, and I can't store the ints into a list.
userEntry = input("Please enter a Fahrenheit temperature: ")
for i in range(4):
userEntry += " " + input("Please enter another fahrenheit:")
Things I AM allowed to use: string methods, index find/search methods, for loops, if statements, while loops.

Something like this will parse the string into space-separated strings, using slices... (I notice the first answer came in while I was working on this, but this is slightly different, so...)
def extractor(mystr):
start = 0
for a in range(len(mystr)):
if mystr[a] == ' ' or mystr[a] == len(mystr) - 1:
temp = mystr[start:a]
print(temp)
start = a + 1
This is more like a C approach, very un-Pythonic, but standard programming fare. If you will only ever have 5 user entries, this is perhaps manageable. If you can't use a list of those variables, or if you have an unknown number of user entries, or if you have to check to make sure the user actually entered a digit and not a letter, then more work is required, but that's the basic C-string parser. Useful to know if you ever want to dive into Python internals I suppose.
If you need to convert each extracted string to an int, and exceptions are allowed, place this inside the if statement to check for type correctness:
try:
myvar1 = int(temp)
except ValueError:
print("Not an int")
Note that if you absolutely cannot use lists, (*or exec as in the above answer) then the only likely option is to keep slicing off the end of the string, i.e you'd have to do something like the following at the end of each if statement, then write that for loop out 4 more times, changing the variable name each time manually.
mystr = mystr[start:len(mystr)]
break
This will of course not work if you have a variable number of user entries. And is incredibly tedious... I suspect the instructor may have intended something different. Note that the real-world process for all that is just:
result = [int(x) for x in mystr.split(' ') if x.isdigit()]

I am not sure what your use case is, and I can not think of a way where you can assign the numbers to variable in a loop, which is what you have to do if you are not allowed to use a loop. The only way I can think of is exec and I do not feel that is allowed for your task. Regardless, I am posting the answer, in case it is usable:
last_space_index = 0
characters_checked = 0
var_num = 1
userEntry = "12.8 -15.8 125.9 0 -40.0"
for character in userEntry:
characters_checked += 1
if character == ' ':
number = float(userEntry[last_space_index:characters_checked])
var_name = 'var'+str(var_num)
var_num += 1
expression = var_name + ' = number'
# expression becomes 'var1 = number'
exec(expression)
last_space_index = characters_checked
last_number = float(userEntry[last_space_index:])
var_name = 'var'+str(var_num)
expression = var_name + ' = last_number'
exec(expression)
# if you know the number of variables you are going to get
print(var1, var2, var3, var4, var5)
# else:
# for i in range(1,var_num+1):
# var_name = 'var'+str(i)
# command = 'print('+var_name+')'
# exec(command)
Output:
>>> 12.8 -15.8 125.9 0 -40.0
You can replace print with whatever you actually want to do.
And this is completely futile if you are allowed to use dictionary, sets or tuple.

How to use raw_input with no input

I want to use raw_input() function in Python.
I want to receive a number from user about the size of the storage
i wrote this down :
number=raw_input()
if the user doesn't provide an input then number = 10 so
if number is None:
number = 10
when i print number, i get nothing i even tried:
if number==-1:
number=10
print"the storage size was set to:",number
The output was:
>
the storage size was set to -1
and not 10
So how should I solve this ?

If you don't care to distinguish between "no input" and "invalid input" (like a non-integer literal), set the default, then attempt to replace it with the user input.
number = 10
try:
number = int(raw_input())
except (EOFError, ValueError):
pass
ValueError will be raised on invalid inputs, including the empty string. EOFError is raised if the user does something like type Control-d in a terminal that interprets that as closing standard input.

First of all you have to convert the input (default for raw_input is a string) into an int using int() function. But be sure that you first check if user typed something. Otherwise you can't convert an empty string. For example:
num_input = raw_input()
if num_input:
number = int(num_input)
Then already the second part of your question should work:
if number == -1:
number = 10
print "the storage size was set to:", number
The second point is that an empty string is not equal to None. None is the only value of NoneType and "" is a string.
So you can compare the input with an empty string, but you can do better (an empty string is evaluated as False):
if not num_input:
number = 10
and to be even more efficient you can simply add an else statement to my first piece of code:
num_input = raw_input()
if num_input:
number = int(num_input)
else:
number = 10

compare number with empty string; not with None .
if number == '':
number = 10

In Python when the variable is empty then it's have inside empty '',
so if you want to check if your variable is unset you need to compare it to '' and not to None.
if number=='':
number=10

You should just compare number and empty.:
if number=="":
number==10

While loop to identify integer in Python

I'm trying to write a program to calculate densities, and I have tried to create a while loop the prevents the user from entering nothing or a non-number for the volume.
But when I run the program the it just loops "You have to type a value" forever. I've tried the same code in a for loop and it does work after inputing 2 numbers.
def GetVolume():
print("How many cublic cm of water does the item displace")
Volume = input()
while Volume == ("") or type(Volume) != int:
print("You have to type a value")
Volume = input()
return float(Volume)

This solution is written assuming you are using Python 3. The problem in your code is that you are assuming that if you type in a number, the input method will return a type int. This is incorrect. You will always get a string back from your input.
Furthermore, if you try to cast int around your input to force an int, your code will raise if you enter a string, with:
ValueError: invalid literal for int() with base 10:
So, what I suggest you do to make your implementation easier is to make use of try/exceptinstead to attempt to convert your value to a float. If it does not work, you prompt the user to keep entering a value until they do. Then you simply break your loop and return your number, type casted to a float.
Furthermore, because of the use of the try/except, you no longer need to put in a conditional check in your while loop. You simply can set your loop to while True and then break once you have satisfied your condition in your code.
Observe the code below re-written with what I mentioned above:
def GetVolume():
print("How many cublic cm of water does the item displace")
Volume = input()
while True:
try:
Volume = float(Volume)
break
except:
print("You have to type a value")
Volume = input()
return Volume

def GetVolume():
Volume = input("How many cublic cm of water does the item displace")
while not Volume.replace('.', '').replace(',', '').isdigit():
Volume = input("You have to type a value")
return float(Volume)
x = GetVolume()
print(x)

You have to modify your while because is validating that is str or different than int. An input will always be an str by default unless you modified the type with int() or float() in your case.
You can use 'try' instead to check for this:
while True:
x = input("How many cubic cm of water does the item displace")
try:
x = float(x)
break
except ValueError:
pass
print('out of loop')

get the list and input from one function and run them in different function

i have a programm that generate the list and then i ask them to tell me what they want to do from the menu and this is where my problem start i was able to get the input form the user to different function but when i try to use the if else condition it doesn't check, below are my code
def menu(x,l):
print (x)
if x == 1:
return make_table(l)
if x == 2:
y= input("enter a row (as a number) or a column (as an uppercase letter")
if y in [ "1",'2','3']:
print("Minmum is:",minimum(y,l))
if x== 3:
print ('bye')
def main():
bad_filename = True
l =[]
while bad_filename == True:
try:
filename = input("Enter the filename: ")
fp = open(filename, "r")
for f_line in fp:
f_str=f_line.strip()
f_str=f_str.split(',')
for unit_str in f_str:
unit=float(unit_str)
l.append(unit)
bad_filename = False
except IOError:
print("Error: The file was not found: ", filename)
#print(l)
condition=True
while condition==True:
print('1- open\n','2- maximum')
x=input("Enter the choice")
menu(x,l)
main()
from the bottom function i can get list and i can get the user input and i can get the data and move it in second function but it wont work after that.thank you

I think your problem is simple, and has nothing to do with how you're passing values between functions.
In main, you're reading a value from the user like this:
x=input("Enter the choice")
The input function:
… reads a line from input, converts it to a string (stripping a trailing newline), and returns that.
So, if the user types 1 at the prompt, you get back the string "1".
Now, you pass that value—perfectly correctly—to menu.
In menu, you then try to compare it to various numbers, like this:
if x == 1:
But this will never be true. A string, like "1", is never equal to a number, like 1. They're not even the same kind of value, much less the same value.
So, you need to do one of two things:
Convert the input to an number. For example, change menu(x,l) to menu(int(x), l). OR…
Write menu to expect strings. For example, change if x == 1: to if x == "1":.
You may be wondering why that print (x) didn't help you debug the problem.
print(x) prints out the end-user-friendly string representation of whatever you give it. That is, it automatically calls the str function for you. For debugging purposes, you often want to use repr instead of str, to get the programmer-friendly string representation instead of the end-user-friendly string representation.
For example, print(str("10")) will print out 10—just like print(str(10)), so you can't tell them apart. But print(repr("10")) will print out '10', unlike print(repr(10)), while prints 10, so you can tell them apart. repr can also help you spot things like strings with invisible characters in them, having special "node" objects from a parser instead of just strings, etc.

Python - display different output if the input is a letter or a number

I want to print the result of the equation in my if statement if the input is a digit and print "any thing" if it is a letter.
I tried this code, but it's not working well. What is wrong here?
while 1:
print '\tConvert ciliuse to fehrenhit\n'
temp = input('\nEnter the temp in C \n\t')
f = ((9/5)*temp +32)
if temp.isdigit():
print f
elif temp == "quit" or temp == "q" :
break
elif temp.isalpha() :
print ' hhhhhhh '

You need to go through your code line by line and think about what type you expect each value to be. Python does not automatically convert between, for example, strings and integers, like some languages do, so it's important to keep types in mind.
Let's start with this line:
temp = input('\nEnter the temp in C \n\t')
If you look at the documentation for input(), input() actually calls eval() on what you type in in Python 2.x (which it looks like you're using). That means that it treats what you type in there as code to be evaluated, just the same as if you were typing it in the shell. So if you type 123, it will return an int; if you type 'abc', it will return a str; and if you type abc (and you haven't defined a variable abc), it will give you an error.
If you want to get what the user types in as a string, you should use raw_input() instead.
In the next line:
f = ((9/5)*temp +32)
it looks like you're expecting temp to be a number. But this doesn't make sense. This line gets executed no matter what temp is, and you're expecting both strings containing digits and strings containing letters as input. This line shouldn't go here.
Continuing on:
if temp.isdigit():
isdigit() is a string method, so here you're expecting temp to be a string. This is actually what it should be.
This branch of the if statement is where your equation should go, but for it to work, you will first have to convert temp to an integer, like this:
c = int(temp)
Also, to get your calculation to work out right, you should make the fraction you're multiplying by a floating-point number:
f = ((9/5.0)*c +32)
The rest of your code should be okay if you make the changes above.

A couple of things first - always use raw_input for user input instead of input. input will evaluate code, which is potentially dangerous.
while 1:
print "\tConvert ciliuse to fehrenhit\n"
temp = raw_input("\nEnter the temp in C \n\t")
if temp in ("quit", "q"):
break
try:
f = ((9.0 / 5.0) * float(temp) + 32)
except ValueError:
print "anything"
Instead of using isalpha to check if input is invalid, use a catch clause for ValueError, which is thrown when a non-numerical value is used.

Why isn't it working? Are you getting an error of any kind?
Straight away I can see one problem though. You are doing the calculation before you verify it as a number. Move the calculation to inside the if temp.isdigit().

Take a look at this for some examples:
http://wiki.python.org/moin/Powerful%20Python%20One-Liners
OK, this works. Only problem is when you quit, you get dumped out of the interpreter.
while 1: import sys; temp=raw_input('\nEnter the temp in C \n\t'); temp.isdigit() and sys.stdout.write('%lf' %((9./5)*float(temp)+32)) or temp=='q' and sys.exit(0) or sys.stdout.write(temp)