 Specman e: How to constrain distribution of values inside list of list?
By : user2786837
Date : March 29 2020, 07:55 AM
it fixes the issue You have soft in there twice, though that isn't the problem. You forgot to put a == before select. I'd write it like this: code :
keep for each (elem) in my_list {
soft elem.comparator[0] == select {
30: [-1000 .. -50]; -- Big negative values
40: [-49 ..49]; -- Medium values
30: [50..1000]; -- Big positive values
};
// Same for elem.comparator[1]
};
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 use Streams to generate a new list taking into account all the first list elements
By : Avinash kumar
Date : March 29 2020, 07:55 AM
With these it helps I would like to map a list to another one. The decision is to be made on all the elements of the first list. Here is how it would look like, in crappy code : , For example: code :
class MatchingClass {
public List<TheEnums> getMatchingEnums(List<String> givenParameters) {
List<TheEnums> enumsList = Arrays.asList(TheEnums.values());
return enumsList.stream()
.filter(e -> givenParameters.containsAll(Arrays.asList(e.getParams())))
.collect(Collectors.toList());
}
}
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How to combine elements from the list taking into account index of the list?
By : Bharath Maripelly
Date : March 29 2020, 07:55 AM
wish of those help Ok. You can iterate through the array starting from index 1. Then inside the first loop have a second loop. There you concatenate the strings from the lists and if list has only one element then just concatenate it. code :
List<String>[] tab = new ArrayList[5];
tab[0] = new ArrayList<String>(Arrays.asList("1242"));
tab[1] = new ArrayList<String>(Arrays.asList("London", "Paris"));
tab[2] = new ArrayList<String>(Arrays.asList("England", "France"));
tab[3] = new ArrayList<String>(Arrays.asList("Finance"));
tab[4] = new ArrayList<String>(Arrays.asList("No"));
StringBuffer buffer;
for(int j=1; j<=2; j++){
buffer = new StringBuffer(tab[0].get(0)).append(" ");
for(int i = 1; i < tab.length; i++){
List<String> list = tab[i];
if(list.size() == 1){
buffer.append(list.get(0)).append(" ");
}else{
buffer.append(list.get(j - 1)).append(" ");
}
}
System.out.println(buffer.toString());
}
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 How to find an item in a list that exists as a sub-string in all the items of a list python
By : Nick Jones
Date : March 29 2020, 07:55 AM
hope this fix your issue Break down the problem into pieces. Start off by making a function that checks if one string is in a sequence of strings: code :
def check_one(string, seqs):
for seq in seqs:
if string not in seq:
return False
return True
def check_all(strings, seq):
result = []
for string in strings:
if check_one(string, seq):
result.append(string)
return result
short_str_list = ["aga", "ttt", "aca"]
seq_list = ["atcgcgtacat", "acatcgggattt", "tttacagtgtgtggg"]
print(check_all(short_str_list, seq_list))
# ['aca']
print([x for x in short_str_list if all(x in y for y in seq_list)])
# ['aca']
print(list(filter(lambda x: all(x in y for y in seq_list), short_str_list)))
# ['aca']
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 How to search in a list whether a list of given items exists in the same order
By : Vinicius de Sá
Date : March 29 2020, 07:55 AM
To fix the issue you can do I need to check if a list contains all the items and in the same sequence using LINQ and if match found return the last matching element. , You could do it with Linq. code :
var allWords = new List<string> { "A", "B", "C", "D", "E", "F","D", "E", "E"};
var words = new List<string>() {"D","E" };
var indexOfLastOccuranceOfSequence = Enumerable.Range(0, allWords.Count - words.Count + 1)
.Select(a => allWords.Skip(a).Take(words.Count)).ToList()
.FindLastIndex(a => a.SequenceEqual(words));
var indexOfLastElement = indexOfLastOccuranceOfSequence+ words.Count() - 1;
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code :
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