hello I am using a stream on my app to get some data. at the end of data the stream stop, what i want is to start again every time stream is done
final stream = Stream.periodic(kDuration, (count) => count)
.take(kLocations.length);
stream.listen((value) => newLocationUpdate(value)
I was searching for hours and didnt find a good solution
Calling .take(kLocations.length) will cause the stream to close once that number of elements have been emitted. For example:
final stream = Stream.periodic(kDuration, (count) => count).take(3);
stream.listen(print); // prints 0, 1, 2, then stops
If instead, you want this to repeat (i.e. emit 0, 1, 2, 0, 1, 2, etc), you can use the modulus operator (%):
final stream = Stream.periodic(kDuration, (count) => count % 3);
stream.listen(print); // prints 0, 1, 2, 0, 1, 2, 0, 1, 2, ...
Related
I want to implement add to checkout in which number of items added is displayed. Plus button adds elements in list and minus removes elements from list. Goal is just to display particular items added and its quantity. I have added items in list want to count length of duplicate items. How we can do that in flutter?
here is your solution. [Null Safe]
void main() {
List<int> items = [1, 1, 1, 2, 3, 4, 5, 5];
Map<int, int> count = {};
items.forEach((i) => count[i] = (count[i] ?? 0) + 1);
print(count.toString()); // {1: 3, 2: 1, 3: 1, 4: 1, 5: 2}
}
I have a dataset that resembles the following:
site_id, species
1, spp1
2, spp1
2, spp2
2, spp3
3, spp2
3, spp3
4, spp1
4, spp2
I want to create a table like this:
site_id, spp1, spp2, spp3, spp4
1, 1, 0, 0, 0
2, 1, 1, 1, 0
3, 0, 1, 1, 0
4, 1, 1, 0, 0
This question was asked here, however the issue I face is that my list of species is significantly greater and so creating a massive query listing each species manually would take a significant amount of time. I would therefore like a solution that does not require this and could instead read from the existing species list.
In addition, when playing with that query, the count() function would keep adding so I would end up with values greater than 1 where multiples of the same species were present in a site_id. Ideally I want a binary 1 or 0 output.
I'm getting a byte array like this one:
[60, 2, 0, 0, 0]
In the documentation there is written this:
uint16 -> heartBeatNum;
uint8 -> rawDataFilesNum;
uint8 -> alertNum
uint8 -> fallsNum
I will explain a little about the device so that you understand and then I ask my question.
The bluetooth device sends an object every minute that is called heartbeat. If this is the first time the object is to use the array looks like this:
After first minute:
[1, 0, 0, 0, 0]
After two minute:
[2, 0, 0, 0, 0]
After three minute:
[3, 0, 0, 0, 0]
After for minute:
[4, 0, 0, 0, 0]
...
Now there are more than 12 that have passed and the array is:
[60, 2, 0, 0, 0]
So I try to understand from the documentation the heartbeat count is the first 16 bytes. I can not figure out how to collect the 60's and the 2's to have the exact heartbeat number.
How does this function?
According to my calculation if I do 60 * 12 = 720
So I should have about 700
Can someone enlighten me how to gather the 16 bytes in int?
Say I have an array of Ints and all elements equal zero. It'd look something like this:
let arr: [Int] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
There are 11 elements in this array in total. I want three of the elements in this array to be the number one. I want these one values to be distributed evenly throughout the array so that it looks something like this:
[0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0]
I want to be able to add however many one's and distribute them evenly (or as close to evenly as possible) no matter how many total elements there are. How could I do this?
Note: For anyone wondering why I need this, I have a collection of strings that when joined together make up a large body of text. Think of the zeroes as the pieces of text and think of the ones as advertisements I am adding in between the text. I wanted to distribute these ads as evenly as possible. I figured this would be a simple way of expressing what I needed.
Maybe you can try this.
var arr: [Int] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
let distribution = arr.count / 3 // 3 is the number of 1s
for (index, value) in arr.enumerated() {
arr[index] = (index + 1) % distribution == 0 ? 1 : value
}
print(arr) // [0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0]
Assuming that the value distribution > 1
I've got a list of some integers, e.g. [1, 2, 3, 4, 5, 10]
And I've another integer (N). For example, N = 19.
I want to check if my integer can be represented as a sum of any amount of numbers in my list:
19 = 10 + 5 + 4
or
19 = 10 + 4 + 3 + 2
Every number from the list can be used only once. N can raise up to 2 thousand or more. Size of the list can reach 200 integers.
Is there a good way to solve this problem?
4 years and a half later, this question is answered by Jonathan.
I want to post two implementations (bruteforce and Jonathan's) in Python and their performance comparison.
def check_sum_bruteforce(numbers, n):
# This bruteforce approach can be improved (for some cases) by
# returning True as soon as the needed sum is found;
sums = []
for number in numbers:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
return n in sums
def check_sum_optimized(numbers, n):
sums1, sums2 = [], []
numbers1 = numbers[:len(numbers) // 2]
numbers2 = numbers[len(numbers) // 2:]
for sums, numbers_ in ((sums1, numbers1), (sums2, numbers2)):
for number in numbers_:
for sum_ in sums[:]:
sums.append(sum_ + number)
sums.append(number)
for sum_ in sums1:
if n - sum_ in sums2:
return True
return False
assert check_sum_bruteforce([1, 2, 3, 4, 5, 10], 19)
assert check_sum_optimized([1, 2, 3, 4, 5, 10], 19)
import timeit
print(
"Bruteforce approach (10000 times):",
timeit.timeit(
'check_sum_bruteforce([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
print(
"Optimized approach by Jonathan (10000 times):",
timeit.timeit(
'check_sum_optimized([1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 200)',
number=10000,
globals=globals()
)
)
Output (the float numbers are seconds):
Bruteforce approach (10000 times): 1.830944365834205
Optimized approach by Jonathan (10000 times): 0.34162875449254027
The brute force approach requires generating 2^(array_size)-1 subsets to be summed and compared against target N.
The run time can be dramatically improved by simply splitting the problem in two. Store, in sets, all of the possible sums for one half of the array and the other half separately. It can now be determined by checking for every number n in one set if the complementN-n exists in the other set.
This optimization brings the complexity down to approximately: 2^(array_size/2)-1+2^(array_size/2)-1=2^(array_size/2 + 1)-2
Half of the original.
Here is a c++ implementation using this idea.
#include <bits/stdc++.h>
using namespace std;
bool sum_search(vector<int> myarray, int N) {
//values for splitting the array in two
int right=myarray.size()-1,middle=(myarray.size()-1)/2;
set<int> all_possible_sums1,all_possible_sums2;
//iterate over the first half of the array
for(int i=0;i<middle;i++) {
//buffer set that will hold new possible sums
set<int> buffer_set;
//every value currently in the set is used to make new possible sums
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums1.insert(myarray[i]);
//transfer buffer into the main set
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums1.insert(*set_iterator);
}
//iterator over the second half of the array
for(int i=middle;i<right+1;i++) {
set<int> buffer_set;
for(set<int>::iterator set_iterator=all_possible_sums2.begin();set_iterator!=all_possible_sums2.end();set_iterator++)
buffer_set.insert(myarray[i]+*set_iterator);
all_possible_sums2.insert(myarray[i]);
for(set<int>::iterator set_iterator=buffer_set.begin();set_iterator!=buffer_set.end();set_iterator++)
all_possible_sums2.insert(*set_iterator);
}
//for every element in the first set, check if the the second set has the complemenent to make N
for(set<int>::iterator set_iterator=all_possible_sums1.begin();set_iterator!=all_possible_sums1.end();set_iterator++)
if(all_possible_sums2.find(N-*set_iterator)!=all_possible_sums2.end())
return true;
return false;
}
Ugly and brute force approach:
a = [1, 2, 3, 4, 5, 10]
b = []
a.size.times do |c|
b << a.combination(c).select{|d| d.reduce(&:+) == 19 }
end
puts b.flatten(1).inspect