looping, want to print out information about an array of dictionary objects at the end of the loop
Tag : iphone , By : user121405
Date : March 29 2020, 07:55 AM
With these it helps Try using [NSString stringWithFormat:@"Something about value: %@", dictionaryItem]
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with single how loop to print last 3rd element from given single linked list?
Tag : cpp , By : user181706
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further With single loop how can we print last 3rd element from given single linked list lets take there are 10 node in give linked list and i am supposed to find last 3rd node? below code is for inserting a node at beginning now how can i print last 3rd element with sing loop? please help me , Keep three pointers: node * prevPrev = null;
node * prev = null;
node * current = head;
prevPrev = prev;
prev = current;
current = current->next;
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While loop with try-catch blocks doesn't wait for input in the try block after looping once
Tag : java , By : Singularity
Date : March 29 2020, 07:55 AM
Any of those help The behaviour you're observing stems from the fact that scanner (assuming you're working with java.util.Scanner) only advances if nextInt() is successful. That means that you have to skip over the faulty input (i.e. by using next()) before proceding.
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Print numbers every two blocks in a for loop
Date : March 29 2020, 07:55 AM
To fix this issue Although Nina's answer is better, for completeness, I post a corrected version of the code that you started with: let n = 8;
for (let i=0; i<=n; i+=2) {
for (let j=i; j<i+2 && j <= n; j++) {
console.log(j);
}
console.log('\n');
}
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How to define blocks of elements in a vector to zero and looping over the next blocks?
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have a large vector which I would like to multiply with a matrix. This is easy to do in r. The issue here is that I would like to define certain blocks of elements within the vector to zero, multiply this with the matrix which then creates a vector, and then repeat for the next block of elements leaving the first block with the original values. , First, I define my vector and matrix. # Data definitions
a <- c(1:6)
b <- matrix(1:36, nrow = 6, ncol = 6)
# Define blocks of zeroes in vector
define_block <- function(v, n = 2){
# Check that v is a multiple of n
if(length(v)%%n)warning("Vector not a multiple of block length")
# Define blocks of zeroes
lapply(1:(length(v)/n), function(x)replace(v, ((x-1)*n + 1):(x*n), 0))
}
# Create lists of blocks
block_list <- define_block(a)
# [[1]]
# [1] 0 0 3 4 5 6
#
# [[2]]
# [1] 1 2 0 0 5 6
#
# [[3]]
# [1] 1 2 3 4 0 0
# Run through block list and bind result into a matrix
do.call(rbind, lapply(block_list, function(x)x%*%b))
# [,1] [,2] [,3] [,4] [,5] [,6]
#[1,] 86 194 302 410 518 626
#[2,] 66 150 234 318 402 486
#[3,] 30 90 150 210 270 330
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