Midterm Exam 2 Review Spring 2004: Count 'em up
Comments, answers, and questions go here:
(Link back to Sp2004 Midterm 2 Review)
Here is your answer
a)
def countAs(string):
x=0
for i in string:
if i == "a":
x = x+1
print x
figure out the spacing on your own because I can't figure out how to do it here and i definately am not reading the FAO's page.
b)
def countAs(string):
x=0
y=0
z=0
for i in string:
if i == "a":
x = x+1
if i == "b":
y = y+1
if i == "c":
z = z+1
print x
print y
print z
the only problem is it doesn't say which count is which. Post that up if you can figure it out.
try print x,y,z
all on the same line
i'm having the same problem figuring out how to say which count is which. Can you help me?
| print "A's",x,"B's",y,"C's",z. Also look at count in the book. Mark Guzdial |
def countLetters(string):
x=0
y=0
z=0
for i in string:
if i == "a":
x = x+1
if i == "b":
y = y+1
if i == "c":
z = z+1
print x,"As",y, "Bs",z, "Cs"
Susan Kommeth
Where is this in the book or the power point slides, because I can't seem to find any examples of this? poof #2
What about calculating percentages of a's, b's, etc???
for some reason i'm having problems just loading the basics here..def countAs(string):
x=0
for i in string:
if i== "a" :
x=x+1
print x
it gives me an error saying 'for i in string' has something wrong with it... any ideas please? Sabrina Hassanali
Oh! I get this problem. Is this right: since you are trying to count the numbers of a,b,c's... you use x,y,z! x is equal to the number of a's you have, y is equal to the number of b's you have, and z is equal to the number of c's you have. So, if you said, there are 6 a's, you know that x=6. My only question is why are you doing +1 for x,y,z.... wouldn't this add one extra number to the count?
x='s 0 originally, so you have to add 1
Just remember, it is a for loop.poof #2
for calculating percentage of a's you would do this
def percA(string)
a = 0
num = 0
for i in string:
if i =="a":
a = a + 1
num = num + 1
percentage = (x/num)*100
print percentage
Laura Bosworth
| I don't think that'll work, Laura, unless you multiply x or num by 1.0. Otherwise, the percentage will always be zero. Mark Guzdial |
Mark could you post the percent example that you did in Breakout today...similar to:
"abracadabra".count("a") ?
thanks
def percents(string):
print "A's",string.count("a")/(1.0 * len(string))
print "B's",string.count("b")/(1.0 * len(string))
print "C's",string.count("c")/(1.0 * len(string))
When i ran the above code, I got 0's...that can't be right...right?
print ("Number of A", x)
print ("Number of B", y)
print ("Number of C", z)
this will say Number of letter and then the actual count.
Lee Yu
def percents(string):
print "A's",string.count("a")
print "B's",string.count("b")
print "C's",string.count("c")
this one took a while. the percentages were tough but only because i left out the 1.0 length, but what is posted should work.
def countAs(string):
a = 0
b = 0
c = 0
for x in string:
if x == "a"
a = a + 1
if x == "b"
b = b + 1
if x == "c"
c = c + 1
length = 1.0len(string)
print a/length, "%A's"
print b/length, "%B's"
print c/length, "%C's"
your code does not work
Hey everyone, I was a tad confused why we all started in on this percentage stuff....if you plug this code into JES, you get the right thing for "Count 'em up" (b)
def countABCs(string):
x=0
y=0
z=0
for i in string:
if i == "a":
x = x+1
if i == "b":
y = y+1
if i == "c":
z = z+1
print x,"As",y, "Bs",z, "Cs"
Christina Sedor
| See the comments page, Christina – I challenged everyone to do these as percentages. Mark Guzdial |
In't this a simpler way to do the first part of this question?
def countAs(input)
new=input.split("")
print new.count("a")
Maybe this doesn't work, but Prof. Guzdial suggested looking at the count method (pg 219 in the book).
Jonathan Laing
| You don't need the split, Jonathan. Strings understand count. Mark Guzdial |
Thanks.
So, it would just be this, correct?:
def countAs(input)
print input.count("a")
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