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Sum number of occurences of string per row


Sum number of occurences of string per row

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Sum number of occurences of string per row
Tag : python , By : Hans-Inge
Date : November 24 2020, 03:41 PM

will help you I have been trying to search an answer for this and I can't find one - I must be misunderstanding something. , try this,
df1['total']= df1.eq('True').sum(axis=1)
df1['total']= df1.eq(True).sum(axis=1)
df1['total']= df1.sum(axis=1)
   score score2  total
0   True  False      1
1   True   True      2
2  False   True      1

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Replace occurences of string that contains any number


Tag : ios , By : chorn
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Regular expressions would do it.
Take a look at NSRegularExpression (iOS >= 4) and this page for how regular expressions work.
// Create your expression
NSError *error = nil;
NSRegularExpression *regex = 
  [NSRegularExpression regularExpressionWithPattern:@"\\b\f[0-9]*\\b"
                                            options:NSRegularExpressionCaseInsensitive
                                              error:&error];

// Replace the matches
NSString *modifiedString = [regex stringByReplacingMatchesInString:string
                                                       options:0
                                                         range:NSMakeRange(0, [string length])
                                                  withTemplate:@"replacement string"];

Split a string on the first number occurences


Tag : javascript , By : Nick Coats
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I didn't get a suitable regex that enable to split such string: , Use:
var result=str.replace(/ (\d+)/gm, ", $1");
 / (\d+)/gm
, $1

Finding the number of occurences of a sub-string in a string without using library functions


Tag : python , By : user123284
Date : March 29 2020, 07:55 AM
I hope this helps you . As you're allowed to use slicing, so you can use that to check whether the character before/after the substring is a space or empty string, if it is then increment count by 1. Note that slices never raise exception, even for out of range indices.
def sub_str_count(s, sub_str):
    le = len(sub_str)
    count = 0
    for i in range(len(s)):
        if s[i:i+le] == sub_str and s[i-1:i] in ('', ' ') and \
                                                       s[i+le:i+le+1] in ('', ' '):
            count += 1
    return count
def check(s, ind):
    """
    Check whether the item present at this index is a space or not.
    For out of bound indices return True.
    For negative indices return True.
    """

    if ind < 0:
        return True
    try:
        return s[ind] == ' '
    except IndexError:
        return True

def sub_str_count(s, sub_str):
    le = len(sub_str)
    count = 0
    for i in range(len(s)):
        if s[i:i+le] == sub_str and check(s, i-1) and check(s, i+le):
            count += 1
    return count

extract number of occurences of `!?` in a string


Tag : regex , By : bdurbin
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Let's say I have this string: S=hello !? how ?! are you ?!?! fine!?!?! thanks !?!?!? bye ?!?!?! bye ! , the regex
\s(!\?)\s
str = 'hello !? how ?! are you ?!?! fine!?!?! thanks !?!?!? bye ?!?!?! bye !';
expression = '\s(!\?)\s';
matchStr = regexp(str, expression, 'match')
length(matchStr)

How to count occurences of number in string?


Tag : javascript , By : Mario Tristan
Date : March 29 2020, 07:55 AM
should help you out You can use reduce and Map
split string on , loop through array and count occurrence of each number Loop through the entries of mapper and sort them if needed in ascending order, if you need in the same order as in input than don't sort it map though the entries to form `number(repetition) form join them by ,
let str ="1,2,3,3,3,4,5,1,4,3,5,2"

function counter(str){
  let mapper = str.split(',').reduce((op,inp)=>{
    op.set(inp, (op.get(inp) || 0 ) + 1)
    return op
  },new Map())

  return [...mapper]
         .sort((a,b)=>a[0]-b[0])  // if you need in always in ascending order
         .map(([key,value])=>`${key}(${value})`)
         .join(',')
}

console.log(counter(str))
console.log(counter("1,2,3,2,3,43,4,53,5,3"))
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