Find repeated elements in a having count

Find repeated elements in a having count

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Find repeated elements in a having count
Tag : sql , By : Valentine
Date : November 29 2020, 04:01 AM

seems to work fine I have a having count, where I detect elements that repeat more than once. , Group by Product as well and it will work:
select MstDocCabId, count (*) as 'Repetition Number'
from skmov (nolock)
group by MstDocCabId, Product
having count(MstDocCabId) > 1

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Find the highest repeated character in string and the count of the repeated character

Tag : ios , By : Guid
Date : March 29 2020, 07:55 AM
Does that help Because the string may have more than a char have same most repeat count, so here is my solution:
- (NSArray *)mostCharInString:(NSString *)string count:(int *)count{
    NSMutableDictionary *dict = [NSMutableDictionary dictionary];
    int len = string.length;
    NSRange theRange = {0, 1};
    for (NSInteger i = 0; i < len; i++) {
        theRange.location = i;
        NSString *charStr = [string substringWithRange:theRange];
        int preCount = 0;
        if ([dict objectForKey:charStr]) {
            preCount = [[dict objectForKey:charStr] unsignedIntegerValue];
        [dict setObject:@(preCount+1) forKey:charStr];
    NSArray *sortValues = [[dict allValues] sortedArrayUsingSelector:@selector(compare:)];
    *count = [[sortValues lastObject] unsignedIntegerValue];
    return  [dict allKeysForObject:@(*count)];
int mostRepeatCount = 0;
NSArray *mostChars = nil;
mostChars = [self mostCharInString:@"aaabbbcccc" count:&mostRepeatCount];
NSLog(@"count:%d char:%@", mostRepeatCount, mostChars);
count:4 char:(
mostChars = [self mostCharInString:@"aaabbbccccdddd" count:&mostRepeatCount];
count:4 char:(

counting matching elements of two vectors but not including repeated elements in the count

Tag : r , By : jbcrail
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Use R's vectorization to your advantage here. There's no looping necessary.
You could use a table to look at the frequencies,
table(z[z %in% x])
# 1 2 4 
# 2 1 1 
sum(table(z[z %in% x]))
# [1] 4

Python Program to Find count of repeated adjacent elements in array

Tag : python , By : browe
Date : March 29 2020, 07:55 AM
this will help Because you have alist[i+1] in for loop , you cant go through list until end , so you have to end it one element before end:
for i in range(len(alist)-1):
    if(alist[i] == alist[i+1]):
        counter+=2    # since you count both elements

alist=[1,2,2,3,3,5,6] # 4
alist=[4,5,6,6,8,8,8,8] # 8

Find repeated elements and it's count from multiple Flux's

Tag : development , By : Elwillow
Date : March 29 2020, 07:55 AM
will be helpful for those in need Since Map is a single item you can generate a Mono> as follows:
Mono<Map<String, Long>> result = Flux.merge(flux1, flux2).groupBy(it -> it).flatMap(group -> Mono.zip(Mono.just(group.key()), group.count())).collectMap(Tuple2::getT1, Tuple2::getT2);
Flux<Tuple2<String, Long>> result = Flux.merge(flux1, flux2).groupBy(it -> it).flatMap(group -> Mono.zip(Mono.just(group.key()), group.count()));

count repeated elements in an array

Tag : c , By : user181445
Date : March 29 2020, 07:55 AM
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