Converting input containing special character to float
$begingroup$
This is the beginning of my program that calculates simple interest. Interest rate will have the following format : 0.97 , 0.67 , 0.17 etc. They won't be bigger than 1. So if the user enter 9 for the interest, program will convert it to 0.09 (by dividing it by 100) . Also user can enter input using '/'. So program will convert input like 97/100 to 0.97.
I wrote the code below. It works but it seems to me that there might be a easier and more elegant solution to this. Maybe using more build-in functions etc. If you help me with that I would be very appreciated.
def toNum(interest):
if '/' not in interest:
if float(interest) > 1:
return float(interest)/100
else:
return float(interest)
else:
l=
n = 0
count = 1
list_interest=
for e in interest:
list_interest.append(e)
for e in list_interest:
if count == 1 or count == 3:
l.append(e)
count = count +1
continue
if e == '/':
n = n + 1
count = count +1
else:
l[n] = l[n] + e
return int(l[0]) / int(l[1])
interest = input("Interest rate: ")
interest = toNum(interest)
print(interest)
python python-3.x
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
$endgroup$
add a comment |
$begingroup$
This is the beginning of my program that calculates simple interest. Interest rate will have the following format : 0.97 , 0.67 , 0.17 etc. They won't be bigger than 1. So if the user enter 9 for the interest, program will convert it to 0.09 (by dividing it by 100) . Also user can enter input using '/'. So program will convert input like 97/100 to 0.97.
I wrote the code below. It works but it seems to me that there might be a easier and more elegant solution to this. Maybe using more build-in functions etc. If you help me with that I would be very appreciated.
def toNum(interest):
if '/' not in interest:
if float(interest) > 1:
return float(interest)/100
else:
return float(interest)
else:
l=
n = 0
count = 1
list_interest=
for e in interest:
list_interest.append(e)
for e in list_interest:
if count == 1 or count == 3:
l.append(e)
count = count +1
continue
if e == '/':
n = n + 1
count = count +1
else:
l[n] = l[n] + e
return int(l[0]) / int(l[1])
interest = input("Interest rate: ")
interest = toNum(interest)
print(interest)
python python-3.x
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
$endgroup$
add a comment |
$begingroup$
This is the beginning of my program that calculates simple interest. Interest rate will have the following format : 0.97 , 0.67 , 0.17 etc. They won't be bigger than 1. So if the user enter 9 for the interest, program will convert it to 0.09 (by dividing it by 100) . Also user can enter input using '/'. So program will convert input like 97/100 to 0.97.
I wrote the code below. It works but it seems to me that there might be a easier and more elegant solution to this. Maybe using more build-in functions etc. If you help me with that I would be very appreciated.
def toNum(interest):
if '/' not in interest:
if float(interest) > 1:
return float(interest)/100
else:
return float(interest)
else:
l=
n = 0
count = 1
list_interest=
for e in interest:
list_interest.append(e)
for e in list_interest:
if count == 1 or count == 3:
l.append(e)
count = count +1
continue
if e == '/':
n = n + 1
count = count +1
else:
l[n] = l[n] + e
return int(l[0]) / int(l[1])
interest = input("Interest rate: ")
interest = toNum(interest)
print(interest)
python python-3.x
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
$endgroup$
This is the beginning of my program that calculates simple interest. Interest rate will have the following format : 0.97 , 0.67 , 0.17 etc. They won't be bigger than 1. So if the user enter 9 for the interest, program will convert it to 0.09 (by dividing it by 100) . Also user can enter input using '/'. So program will convert input like 97/100 to 0.97.
I wrote the code below. It works but it seems to me that there might be a easier and more elegant solution to this. Maybe using more build-in functions etc. If you help me with that I would be very appreciated.
def toNum(interest):
if '/' not in interest:
if float(interest) > 1:
return float(interest)/100
else:
return float(interest)
else:
l=
n = 0
count = 1
list_interest=
for e in interest:
list_interest.append(e)
for e in list_interest:
if count == 1 or count == 3:
l.append(e)
count = count +1
continue
if e == '/':
n = n + 1
count = count +1
else:
l[n] = l[n] + e
return int(l[0]) / int(l[1])
interest = input("Interest rate: ")
interest = toNum(interest)
print(interest)
python python-3.x
python python-3.x
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
asked Nov 26 '18 at 13:51
ikadorusikadorus
234
234
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For reading a fraction such as "97/100", you can use the fractions library.
For example:
from fractions import Fraction
f = Fraction("97/100")
print(float(f)) # prints 0.97
And because the constructor also takes a float, we can remove the check for /. Therefore, the final code is:
from fractions import Fraction
def toNum(interest):
f = Fraction(interest)
f = float(f)
if f > 1:
f /= 100
return f
print(toNum("97/100")) # prints 0.97
print(toNum(0.97)) # prints 0.97
print(toNum(9)) # prints 0.09
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
$endgroup$
1
$begingroup$
Why convert tofloatat all and not keep afractions.Fractionobject all along?
$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
add a comment |
$begingroup$
While the answer by @esote is correct and I would also recommend using the fractions module, you should also work on your text parsing. In this case you could have used a simple str.split and map to parse the string containing a /:
if "/" in interest:
numerator, denominator = map(int, interest.split("/"))
return numerator / denominator
Note that int ignores whitespace, so this works with both "97/100", "97 / 100" and any combination thereof.
Note also that using sensible names makes it immediately obvious what this code does.
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For reading a fraction such as "97/100", you can use the fractions library.
For example:
from fractions import Fraction
f = Fraction("97/100")
print(float(f)) # prints 0.97
And because the constructor also takes a float, we can remove the check for /. Therefore, the final code is:
from fractions import Fraction
def toNum(interest):
f = Fraction(interest)
f = float(f)
if f > 1:
f /= 100
return f
print(toNum("97/100")) # prints 0.97
print(toNum(0.97)) # prints 0.97
print(toNum(9)) # prints 0.09
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
$endgroup$
1
$begingroup$
Why convert tofloatat all and not keep afractions.Fractionobject all along?
$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
add a comment |
$begingroup$
For reading a fraction such as "97/100", you can use the fractions library.
For example:
from fractions import Fraction
f = Fraction("97/100")
print(float(f)) # prints 0.97
And because the constructor also takes a float, we can remove the check for /. Therefore, the final code is:
from fractions import Fraction
def toNum(interest):
f = Fraction(interest)
f = float(f)
if f > 1:
f /= 100
return f
print(toNum("97/100")) # prints 0.97
print(toNum(0.97)) # prints 0.97
print(toNum(9)) # prints 0.09
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
$endgroup$
1
$begingroup$
Why convert tofloatat all and not keep afractions.Fractionobject all along?
$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
add a comment |
$begingroup$
For reading a fraction such as "97/100", you can use the fractions library.
For example:
from fractions import Fraction
f = Fraction("97/100")
print(float(f)) # prints 0.97
And because the constructor also takes a float, we can remove the check for /. Therefore, the final code is:
from fractions import Fraction
def toNum(interest):
f = Fraction(interest)
f = float(f)
if f > 1:
f /= 100
return f
print(toNum("97/100")) # prints 0.97
print(toNum(0.97)) # prints 0.97
print(toNum(9)) # prints 0.09
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
$endgroup$
For reading a fraction such as "97/100", you can use the fractions library.
For example:
from fractions import Fraction
f = Fraction("97/100")
print(float(f)) # prints 0.97
And because the constructor also takes a float, we can remove the check for /. Therefore, the final code is:
from fractions import Fraction
def toNum(interest):
f = Fraction(interest)
f = float(f)
if f > 1:
f /= 100
return f
print(toNum("97/100")) # prints 0.97
print(toNum(0.97)) # prints 0.97
print(toNum(9)) # prints 0.09
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
edited Nov 26 '18 at 15:03
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
answered Nov 26 '18 at 14:11
esoteesote
2,6391934
2,6391934
1
$begingroup$
Why convert tofloatat all and not keep afractions.Fractionobject all along?
$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
add a comment |
1
$begingroup$
Why convert tofloatat all and not keep afractions.Fractionobject all along?
$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
1
1
$begingroup$
Why convert to
float at all and not keep a fractions.Fraction object all along?$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
Why convert to
float at all and not keep a fractions.Fraction object all along?$endgroup$
– Mathias Ettinger
Nov 26 '18 at 15:59
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
$begingroup$
@MathiasEttinger That's also a good option, just depends on preference.
$endgroup$
– esote
Nov 26 '18 at 16:14
add a comment |
$begingroup$
While the answer by @esote is correct and I would also recommend using the fractions module, you should also work on your text parsing. In this case you could have used a simple str.split and map to parse the string containing a /:
if "/" in interest:
numerator, denominator = map(int, interest.split("/"))
return numerator / denominator
Note that int ignores whitespace, so this works with both "97/100", "97 / 100" and any combination thereof.
Note also that using sensible names makes it immediately obvious what this code does.
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
$endgroup$
add a comment |
$begingroup$
While the answer by @esote is correct and I would also recommend using the fractions module, you should also work on your text parsing. In this case you could have used a simple str.split and map to parse the string containing a /:
if "/" in interest:
numerator, denominator = map(int, interest.split("/"))
return numerator / denominator
Note that int ignores whitespace, so this works with both "97/100", "97 / 100" and any combination thereof.
Note also that using sensible names makes it immediately obvious what this code does.
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
$endgroup$
add a comment |
$begingroup$
While the answer by @esote is correct and I would also recommend using the fractions module, you should also work on your text parsing. In this case you could have used a simple str.split and map to parse the string containing a /:
if "/" in interest:
numerator, denominator = map(int, interest.split("/"))
return numerator / denominator
Note that int ignores whitespace, so this works with both "97/100", "97 / 100" and any combination thereof.
Note also that using sensible names makes it immediately obvious what this code does.
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
$endgroup$
While the answer by @esote is correct and I would also recommend using the fractions module, you should also work on your text parsing. In this case you could have used a simple str.split and map to parse the string containing a /:
if "/" in interest:
numerator, denominator = map(int, interest.split("/"))
return numerator / denominator
Note that int ignores whitespace, so this works with both "97/100", "97 / 100" and any combination thereof.
Note also that using sensible names makes it immediately obvious what this code does.
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
answered Nov 27 '18 at 11:27
GraipherGraipher
23.9k53585
23.9k53585
add a comment |
add a comment |
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