Total number count

[Lord Cat]

Hey i'm Looking for a way to be able to put numbers into a list and then get the computer to tell me how many of that number there is

Like say i had a list ('1,1,10,18,34,134,345,1001')
And i wanted to know how many 1's there are in total of every number including '1001'
than it would tell me how many
Like 1's = 7
how would that be done?

In python plz ,I'm a noob to python so i need some help,Ty

[deer dance]

What version are you using?

[PythonPower]

You want something like:

arr = [1, 1, 10, 18, 34, 134, 345, 1001]

def ones(n):
    s = 0
    while n > 0:
        if n%10 == 1: s += 1
        n //= 10
    return s

t = 0
for x in arr: t += ones(x)
print(t)

That's Python 3k code by the way.

[Lord Cat]

Something wrong with it when i put it in...I'm using Python 3.0 too

arr = [1, 1, 10, 18, 34, 134, 345, 1001]

def ones(n):
    s = 0
    while n > 0:
        if n%10 == 1: s += 1
        n //= 10
    return s

t = 0
for x in arr: t += ones(x)
print(t)
SyntaxError: invalid  syntax (<pyshell#26, line 2>)

[PythonPower]

It sure works for me. Try creating a file and pasting the code in; then run it like:

python program.py

Less likely to error than pasting it in the "live command-line".

[kiddies]

nice basic tutorial for python using a function sir....

[Hignar]

You want something like:

arr = [1, 1, 10, 18, 34, 134, 345, 1001]

def ones(n):
    s = 0
    while n > 0:
        if n%10 == 1: s += 1
        n //= 10
    return s

t = 0
for x in arr: t += ones(x)
print(t)

That's Python 3k code by the way.

This doesn't return what Lord Cat wants though. Your code only returns numbers that end in 1. If I understood the original post he wants to count all the 1s including the ones in 134 etc.

Not sure off the top of my head how you could do this. Maybe convert each number to a string and check each digit in the number that way?

[Hignar]

I've been thinking about this and I couldn't come up with a way of testing the integer directly, so ended up converting each one to a string and testing each character of the string for equality to '1'.

arr = [1, 1, 10, 18, 34, 134, 345, 1001]
strarr = []
count = 1

for x in arr:
   strarr.append(str(x)) 

for x in strarr:
   for c in x:
      if c == '1':
         count += 1

print count

I'm new to python so this probably isn't the easiest or most efficient way of doing this but it gets the right answer.

[CallBackGuy]

I've been thinking about this and I couldn't come up with a way of testing the integer directly, so ended up converting each one to a string and testing each character of the string for equality to '1'.

count = 1

You may want to start at zero, otherwise it will output 8 instead of 7
Here's mine it's very much the same as yours .-.

arr = [1, 1, 10, 18, 34, 134, 345, 1001]
count = 0
for strarr in str(arr):
    for x in strarr:
        if x == '1':
            count+=1
print count

3 lines anyone? xD

arr, count = [1, 1, 10, 18, 34, 134, 345, 1001], []
[ ( lambda x: [ count.append(x) if x == '1' else '' for x in strarr ] )(strarr) for strarr in str( arr ) ]
print len( count )

[Hignar]

You may want to start at zero, otherwise it will output 8 instead of 7

Oops. Too many 1s floating about I guess.

Didn't realise you could convert the arr to strarr like that.

Ah, turns out you can't. What you've done is turn the array into one string ('[1, 1, 10, 18, 34, 134, 345, 1001]') and tested each character. It produces the same result, but I prefer my method as it leaves you with the ability to manipulate each string in the array if you want to. Admittedly, not a concern here though.