[Tutor] Count for loops

[D.V.N.Sarma డి.వి.ఎన్.శర్మ]

I will go for this modification of the original code.

file_path = "C:/Users/Rafael/PythonCode/PiDigits.txt"
with open(file_path) as a:
    b = a.read()

get_year = input("What year were you born? ")

count = 0
b= '3'+b[2:]
n = len(b)
for i in range(n-4):
    if b[i:i+4] == get_year:
        count += 1
print("Your birth date occurs %s times in PI!" % (count))

regards,
Sarma.

On Tue, Apr 4, 2017 at 12:54 AM, Mats Wichmann <mats at wichmann.us> wrote:

> On 04/03/2017 10:16 AM, Alan Gauld via Tutor wrote:
> > On 03/04/17 16:42, D.V.N.Sarma డి.వి.ఎన్.శర్మ wrote:
> >> Sorry. That was  stupid of me. The loop does nothing.
> >
> > Let me rewrite the code with some different variable names...
> >
> >>>> with open(file_path) as a:
> >>>>     b = a.read()
> >
> > with open (file_path) as PI_text:
> >      PI_as_a_long_string = PI_text.read()
> >
> >>>> for year in b:
> >
> > for each_char in PI_as_a_long_string:
> >
> >>>>     if get_year in b:
> >>>>         count += 1
> >
> >       if get_year in PI_as_a_long_string:
> >           count += 1
> >
> >>>> print("Your birth date occurs %s times in PI!" % (count))
> >
> > if count:
> >    print("Your birth date occurs in PI, I checked, count, "times!")
> >
> >
> > Aren't meaningful variable names great? We should all do
> > them more often.
>
>
> So the takeaways here are:
>
> in the first ("non-counting") sample, there's no need to use a loop,
> because you're going to quit after the outcome "in" or "not in" right
> away - there's no loop behavior at all.
>
> for year in b:
>     if get_year in b:
>         print("Your year of birth occurs in PI!")
>         break
>     else:
>         print("Your year of birth does not occur in PI.")
>         break
>
> for the counting version you could do something that searches for the
> year repeatedly, avoiding starting over from the beginning so you're
> actually finding fresh instances, not the same one (hint, hint). There
> are several ways to do this, including slicing, indexing, etc.  Alan's
> suggestion looks as good as any.
>
> Or, if you don't care about overlapping cases, the count method of a
> string will do just fine:
>
> PI_as_a_long_string.count(year)
>
> If you're talking about 4-digit year numbers using a Western calendar in
> digits of PI, the overlap effect seems unlikely to matter - let's say
> the year is 1919, do we think PI contains the sequence 191919? count
> would report back one instead of two in that case. In other cases it
> might matter; count is written specifically to not care about overlaps:
> "Return the number of (non-overlapping) occurrences"  So that's worth
> keeping in mind when you think about what you need from
> substrings-in-strings cases.
>
>
>
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