beginner here so be kind.
This is my example code.
def iq_test(numbers): for i in numbers: if int(i) %2 != 0: return i
This bit of code always returns the invalid literal for base10... which is weird as when the mod operator is == 0, it doesn't return the error. Where am I going wrong?
I've tried .isdigit, tried to use a float instead but no joy.
You haven't stated what your requirement is, so I will guess that it is to "extract the first odd number from a string containing a mixture of numbers and letters".
You also wonder why i % 2 == 0 allows some specific test to work, but that might only be because the string being tested only contains even numbers.
Here is your function adjusted to answer that hypothetical question:
def iq_test(numbers):
for i in numbers:
if str(i).isdigit() and int(i) %2 != 0:
return i
Note the stringify requirement for testing each element of the string/list as integer.
Related
I am having some issues with some code I wrote for this problem:
“Write a function namedd calc that will evaluate a simple arithmetic expression. The input to your program will be a string of the form:
operand1 operator operand2
where operand1 and operand2 are non-negative integers and operator is a single-character operator, which is either +, -, or *. You may assume that there is a space between each operand and the operator. You may further assume that the input is a valid mathemat- ical expression, i.e. your program is not responsible for the case where the user enters gibberish.
Your function will return an integer, such that the returned value is equal to the value produced by applying the given operation to the given operands.
Sample execution:
calc("5 + 10") # 15
“You may not use the split or eval functions in your solution.
Hint: the hard part here is breaking the input string into its three component. You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters. Be careful of off-by-one errors in using the slice operator.
Hint: it’s best to test your code as you work. The first step should be to break the input string into its three components. Write a program that does that, have it print out the operator and the two operands on separate lines, and test it until you are convinced that it works. Then, modifying it to perform the desired mathematical operation should be straightforward. Test your program with several different inputs to make sure it works as you expect.”
Here is my code:
def calc(exp):
operand1 = int(exp[:1])
operand2 = int(exp[4:6])
operator = exp[2:3]
if(operator == "+"):
addition = operand1+operand2
return addition
if(operator == "-"):
subtraction = operand1-operand2
return subtraction
if(operator == "*"):
multiplication = operand1*operand2
return multiplication
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
My code does not fully meet the criteria of this question. It only works for single digit numbers. How can I make my code work for any number?
Like:
“504 + 507”
”5678 + 76890”
and so on?
Thank you. Any help is appreciated.
As the hint says, get the position of the first and last space of the expression, use it to extract the operand and the operators, and then evaluate accordingly.
def calc(exp):
#Get the position for first space with find
low_idx = exp.find(' ')
#Get the position for last space with rfind
high_idx = exp.rfind(' ')
#Extract operators and operand with slice, converting operands to int
operand1 = int(exp[0:low_idx])
operator = exp[low_idx+1:high_idx]
operand2 = int(exp[high_idx:])
result = 0
#Evaluate based on operator
if operator == '+':
result = operand1 + operand2
elif operator == '-':
result = operand1 - operand2
elif operator == '*':
result = operand1 * operand2
return result
print(calc("5 + 10"))
print(calc("4 - 8"))
print(calc("4 * 3"))
print(calc("504 + 507"))
print(calc("5678 + 76890"))
#15
#-4
#12
#1011
#82568
The answer is in the specification:
You may use the find and rfind functions to find the position of the first and last space, and then use the slice operator (that is, s[startindex:endindex]) to extract the relevant range of characters.
find and rfind are methods of string objects.
You could split it into three components using this code: (note: this doesn't use split or eval)
def splitExpression(e):
numbers = ["1","2","3","4","5","6","7","8","9","0"] # list of all numbers
operations = ["+","-","*","/"] # list of all operations
output = [] # output components
currentlyParsing = "number" # the component we're currently parsing
buildstring = "" # temporary variable
for c in e:
if c == " ":
continue # ignore whitespace
if currentlyParsing == "number": # we are currently parsing a number
if c in numbers:
buildstring += c # this is a number, continue
elif c in operations:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "operation" # we are expecting an operation now
else:
pass # unknown symbol!
elif currentlyParsing == "operation": # we are currently parsing an operation
if c in operations:
buildstring += c # this is an operation, continue
elif c in numbers:
output.append(buildstring) # this component has reached it's end
buildstring = c
currentlyParsing = "number" # we are expecting a number now
else:
pass # unknown symbol!
if buildstring: # anything left in the buffer?
output.append(buildstring)
buildstring = ""
return output
Usage: splitExpression("281*14") returns ["281","*","14"]
This function also accepts spaces between numbers and operations
You can simply take the string and use the split method for the string object, which will return a list of strings based on some separator.
For example:
stringList = "504 + 507".split(" ")
stringList will now be a list such as ["504", "+", "507"] due to the separator " " which is a whitespace. Then just use stringList[1] with your conditionals to solve the problem. Additionally, you can use int(stringList[0]) and int(stringList[2]) to convert the strings to int objects.
EDIT:
Now I realized that your problem said to use find() instead of split(). Simply use the logic above but instead find(" ") the first whitespace. You will then need to find the second whitespace by slicing past the first whitespace using the two additional arguments available for find().
You need to split the string out instead of hard coding the positions of the indexes.
When coding you want to try to make your code as dynamic as possible, that generally means not hard coding stuff that could be a variable or in this case could be grabbed from the spaces.
Also in the if statements I modified them to elif as it is all one contained statement and thus should be grouped.
def calc(exp):
vals = exp.split(' ')
operand1 = int(vals[0])
operand2 = int(vals[2])
operator = vals[1]
if operator == '+':
return operand1+operand2
elif operator == '-':
return operand1-operand2
else:
return operand1*operand2
I want to have a single assignment variable but make the assignment later (in a different block).
For example:
let n: int
try:
n = parseInt paramStr(1)
except ValueError:
quit "Expected Number
Is this impossible right now? If so, why?
Also what is the best alternative in this situation?
I considered this:
let n =
try:
parseInt paramStr(2)
except ValueError:
-1
if n == -1:
quit "Expected a number"
but doesn't look nice of course...
It is impossible given the current implementation, your alternative seems reasonable to me (or use a var with initial value of -1). Otherwise you can use the parseopt module to abstract the command line input verification logic.
You can do something like:
let n =
try:
var n: int
let processedChars = paramStr(1).parseInt(n)
if processedChars == 0:
quit "Expected number"
n
except ValueError:
quit "Parsed integer is out of the valid range"
Also note that parseInt doesn't return the parsed value, but rather the number of parsed characters.
I am writing #A code that prints hello if 1 is stored in spam, howdy if 2 is stored in spam and Greetings if anything else is stored.
see what I've written
while True:
spam = int(input())
if spam == 1:
print('Hello')
if spam == 2:
print('Howdy')
else:
print('Greeting')
break
Any time I type 2, I get Howdy and a blank space to type in some word figure, but once I hit enter, it gives this error
"ValueError: invalid literal for int() with base 10: ''
also, when I enter 1, it prints the whole instruction in the code.
Please how do i write this code
int(input()) means take input and convert that input to int that means if you type any character other than integer (or a blank space to type in some word figure) you will get following error.
invlaid literal for int() with base 10:
Also when you take 1 as an input first Hello will be printed then Greeting would be printed as you are not using if..elif..else
s = 'eljwboboblejr' # dont paste into grader
count = 0
for i in range (len(s)):
if s[i:i+3]== 'bob':
count+=1
print('Number of times bob occurs is: ' + str(count))
I do not get how len is working here, or if s[i:i+3] == 'bob'
So what happens here is that the i goes through all the letters, and slice all the letters by i and i+3 in each loop. What len is doing is just taking the length of s (basically how many characters there are in it) and returning it as an integer. What the s[i:i+3] == 'bob' is doing is determining if the sliced string is equal to 'bob'. So imagine that the i represents all the letters in the s string. So if the sliced string that is contained by the i and i+3 has 'bob' in it, it returns true. It's not the greatest of explanations, but I hope it helps.
documentation for len is here:
https://docs.python.org/3.2/library/functions.html#len
It will be implemented in string as a magic private function (__len__, I believe).
documentation for range is here:
https://docs.python.org/3.2/library/functions.html#range
With one arg, range generates integers 0 to that arg (excluding arg itself).
The slice in the loop evaluates to 'elj', then 'ljw', then 'jwb', ... in subsequent iterations. The slice [a:b] doesn't include the b'th element.
I was tasked with reversing an integer recursively. I have an idea of how to formulate my base case but I'm unsure of what to put outside of the if statement. The parts I was unsure about are commented with question marks. With the first part, I don't know what to put and with the second part I'm unsure about whether it is correct or not.Thank you for the help.
Note: I'd like to avoid using external functions such as imports and things like these if possible.
def reverseDisplay(number):
if number < 10:
return number
return # ??????????
def main():
number = int(input("Enter a number: "))
print(number,end="") #???????????
reverseDisplay(number)
main()
I'm not going to give you the answer, but I'll give some hints. It looks like you don't want to convert it to a string -- this makes it a more interesting problem, but will result in some funky behavior. For example, reverseDisplay(100) = 1.
However, if you don't yet have a good handle on recursion, I would strongly recommend that you convert the input to a string and try to recursively reverse that string. Once you understand how to do that, an arithmetic approach will be much more straightforward.
Your base case is solid. A digit reversed is that same digit.
def reverseDisplay(n):
if n < 10:
return n
last_digit = # ??? 12345 -> 4
other_digits = # ??? You'll use last_digit for this. 12345 -> 1234
return last_digit * 10 ** ??? + reverseDisplay(???)
# ** is the exponent operator. If the last digit is 5, this is going to be 500...
# how many zeroes do we want? why?
If you don't want to use any string operations whatsoever, you might have to write your own function for getting the number of digits in an integer. Why? Where will you use it?
Imagine that you have a string 12345.
reverseDisplay(12345) is really
5 + reverseDisplay(1234) ->
4 + reverseDisplay(123) ->
3 + reverseDisplay(12) ->
2 + reverseDisplay(1) ->
1
Honestly, it might be a terrible idea, but who knows may be it will help:
Convert it to string.
Reverse the string using the recursion. Basically take char from the back, append to the front.
Parse it again.
Not the best performing solution, but a solution...
Otherwise there is gotta be some formula. For instance here:
https://math.stackexchange.com/questions/323268/formula-to-reverse-digits
Suppose you have a list of digits, that you want to turn into an int:
[1,2,3,4] -> 1234
You do this by 1*10^3 + 2*10^2 + 3*10^1 + 4.*10^0. The powers of 10 are exactly reversed in the case that you want to reverse the number. This is done as follows:
def reverse(n):
if n<10:
return n
return (n%10)*10**(int(math.log(n,10))) + reverse(n//10)
That math.log stuff simply determines the number of digits in the number, and therefore the power of 10 that should be multiplied.
Output:
In [78]: reverse(1234)
Out[78]: 4321
In [79]: reverse(123)
Out[79]: 321
In [80]: reverse(12)
Out[80]: 21
In [81]: reverse(1)
Out[81]: 1
In [82]: reverse(0)
Out[82]: 0
Does exactly what #GregS suggested in his comment. Key to reverse is to extract the last digit using the modulos operator and convert each extracted digit to a string, then simply join them back into the reverse of the string:
def reverseDisplay(number):
if number < 10:
return str(number)
return str(number % 10) + reverseDisplay(number / 10)
def main():
print (reverseDisplay(int(input("Enter a number: "))))
main()
Alternative method without using recursion:
def reverseDisplay(number):
return str(number)[::-1]