I have the following data set:
Word-1-random
Word-2-random
Word-3-random
Word-4-random
upto
Word-19-random
Other-Word-1-random
Other-Word-2-random
Other-Word-3-random
Other-Word-4-random
upto
Other-Word-19-random
Now i want to do a match on a group of date, digits 1 - 5, 6-10, 11-15 etc.
I thought it was the following:
match("^Word%-d[1-5]%-",string)
match("%-Word%-d[1-5]%-",string)
Acording to you example, the idea could be something like this:
<script src="https://github.com/fengari-lua/fengari-web/releases/download/v0.1.4/fengari-web.js"></script>
<script type="application/lua">
local text = [[
Word-1-random
Word-2-random
Word-3-random
Word-4-random
upto
Word-19-random
Other-Word-1-random
Other-Word-2-random
Other-Word-3-random
Other-Word-4-random
Other-Word-5-random
Other-Word-6-random
Other-Word-7-random
Other-Word-8-random
Other-Word-9-random
Other-Word-10-random
Other-Word-11-random
]]
local s1,s2
local grp,cnt = 0,0
text:gsub('(%S+%-)(%d+)(%-%S+)', function(p1,n,p2)
if s1 ~= p1 or s2 ~= p2 or cnt == 5 then
print('Group'..grp)
s1,s2 = p1,p2
cnt = 0
grp = grp + 1
end
print(p1..n..p2)
cnt = cnt + 1
end)
</script>
Related
I have seen different solutions to the same problem, but none of them seem to use the approach I used. So here I'm trying to solve the classical coin-change problem in a bottom up dynamic programming approach using a DP table.
int[][] dp = new int[nums.length+1][target+1];
for (int i = 0; i <= nums.length; ++i)
dp[i][0] = 1;
for (int i = 1; i < dp.length; ++i) {
for (int j = 1; j < dp[i].length; ++j) {
if (nums[i-1] <= j)
dp[i][j] = dp[i-1][j] + dp[i][j-nums[i-1]];
else
dp[i][j] = dp[i-1][j];
}
}
The above code generates the table. For fun if I have: {2,3,5} coins, and want to exchange 8, the table would look like:
1 0 0 0 0 0 0 0 0
1 0 1 0 1 0 1 0 1
1 0 1 1 1 1 2 1 2
1 0 1 1 1 2 2 2 3
For the above the following method seem to be working well:
current <- 4
while (current > 0) do
i <- current
j <- 8
while (j > 0) do
if (dp[i][j] != dp[i-1][j]) then
nums[i-1] is part of the current solution
j <- j-1
else
i <- i-1
end
end
current <- current-1
end
Walking through for the above example, we get the following solutions:
1) [5,3]
2) [3,3,2]
3) [2,2,2,2]
Which is great! At least I thought, until I tried: {1,2} with a T=4. For this the table looks like:
1 0 0 0 0
1 1 1 1 1
1 1 2 2 3
With the above algorithm to print all solutions, I only get:
[2,2]
[1,1,1,1]
Which means I won't recover [2,1,1]. So this question is not about the generic how to print the solutions for different approaches to this problem, but how can I read the above DP table to find all solutions.
Well I have one solution but sure... I can see why the other similar answers are using different approaches for this problem. So the algorithm to generate all the possible answers from the above table looks like:
private List<List<Integer>> readSolutions(int[] nums, int[][] dp, int currentRow, int currentColumn, List<Integer> partialSolution) {
if (currentRow == 0) {
return new ArrayList<>();
}
if (currentColumn == 0) {
return new ArrayList<List<Integer>>(Collections.singleton(partialSolution));
}
List<List<Integer>> solutions = new ArrayList<>();
if (dp[currentRow][currentColumn] != dp[currentRow-1][currentColumn]) {
List<Integer> newSolution = new ArrayList<>(partialSolution);
newSolution.add(nums[currentRow-1]);
solutions.addAll(readSolutions(nums, dp, currentRow, currentColumn-nums[currentRow-1], newSolution));
solutions.addAll(readSolutions(nums, dp, currentRow-1, currentColumn, partialSolution));
return solutions;
}
return readSolutions(nums, dp, currentRow-1, currentColumn, partialSolution);
}
The logic on the other hand is simple. Take the above table for example:
0 1 2 3 4
0 1 0 0 0 0
1 1 1 1 1 1
2 1 1 2 2 3
To get a single solution we start from the bottom right corner. If the value does match with the value directly above us, we move up. If it doesn't we move left by the amount corresponding to the row we're in. To generate all answers on on the other hand from the above table...
we're at some position (i,j)
if the value at (i-1,j) is the same as (i,j) we make a recursive call to (i-1,j)
if the values do not match, we have 2 choices...
we can use the number corresponding to the current row, and recurse into (i,j-n)
we can skip the number and check if we can create (i,j) instead by using a recursive call to (i-1,j) instead.
if we reach the first row, we return an empty list
if we reach the first column, we return what we have already found, which will have the sum of target.
Looks awful right, but works as expected.
I have a 5x-5 maze specified as follows.
r = [1 0 1 1 1
1 1 1 0 1
0 1 0 0 1
1 1 1 0 1
1 0 1 0 1];
Where 1's are the paths and 0's are the walls.
Assume I have a function foo(policy_vector, r) that maps the elements of the policy vector to the elements in r. For example 1=UP, 2=Right, 3=Down, 4=Left. The MDP is set up such that the wall states are never realized so policies for those states are ignored in the plot.
policy_vector' = [3 2 2 2 3 2 2 1 2 3 1 1 1 2 3 2 1 4 2 3 1 1 1 2 2]
symbols' = [v > > > v > > ^ > v ^ ^ ^ > v > ^ < > v ^ ^ ^ > >]
I am trying to display my policy decision for a Markov Decision Process in the context of solving a maze. How would I plot something that looks like this? Matlab is preferable but Python is fine.
Even if some body could show me how to make a plot like this I would be able to figure it out from there.
function[] = policy_plot(policy,r)
[row,col] = size(r);
symbols = {'^', '>', 'v', '<'};
policy_symbolic = get_policy_symbols(policy, symbols);
figure()
hold on
axis([0, row, 0, col])
grid on
cnt = 1;
fill([0,0,col,col],[row,0,0,row],'k')
for rr = row:-1:1
for cc = 1:col
if r(row+1 - rr,cc) ~= 0 && ~(row == row+1 - rr && col == cc)
fill([cc-1,cc-1,cc,cc],[rr,rr-1,rr-1,rr],'g')
text(cc - 0.55,rr - 0.5,policy_symbolic{cnt})
end
cnt = cnt + 1;
end
end
fill([cc-1,cc-1,cc,cc],[rr,rr-1,rr-1,rr],'b')
text(cc - 0.70,rr - 0.5,'Goal')
function [policy_symbolic] = get_policy_symbols(policy, symbols)
policy_symbolic = cell(size(policy));
for ii = 1:length(policy)
policy_symbolic{ii} = symbols{policy(ii)};
end
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So, I have two strings that are each a max length of 100.
Dim a as String ''has a max length of 100
Dim b as String ''has a max length of 100
These two strings need to be truncated and combined into a new string.
Dim c as String 'has a max length of 100
I need to be able to truncate each string appropriately so that I can get string c as close to 100. I was going to do a bunch of statements by 25 to truncate each one.
if a.length = 100 and b.length =0 then
return a
else if a.length = 100 andalso b.length <= 25 then
return a.truncate(75) & b
else if a.length = 100 andalso b.length <= 50 then
return a.truncate(50) & b
else if....
and so one to hit all the scenarios...
I feel like there is a better way to do this and a more efficient way so that i may not hit scenarios like a.length = 100 and b.length = 51. I would be truncating more characters then needed.
Any suggestions?? Please critique me as needed.
EDIT, This is vb.Net..not C# (I'm between Projects) Sorry!
The reason i do not want to just add them together and truncate them is because if both strings are 100 in length, it will completely truncate off the second string. If they are both 100 then I would want to truncate string a to 50 in length and string b to 50 in length so when they are combined they are 100 total. In other words I need some text from both strings.
If the total length of the strings is greater than the limit then you could take a fraction of each in proportion to their lengths:
Module Module1
Function CombineWithLengthConstraint(a As String, b As String, totalLength As Integer) As String
' trivial case 1:
If totalLength < 1 Then
Return String.Empty
End If
Dim aLen = Len(a)
Dim bLen = Len(b)
' trivial case 2:
If aLen + bLen <= totalLength Then
Return a & b
End If
' impossible-to-satisfy-equably case:
If totalLength = 1 Then
If aLen > 0 Then
Return a.Substring(0, 1)
ElseIf bLen > 0 Then
Return b.Substring(0, 1)
Else
Return String.Empty
End If
End If
' aportion the lengths of the strings to be taken in the ratio of their lengths:
Dim aFrac = CInt(Math.Round(aLen / (aLen + bLen) * totalLength, MidpointRounding.AwayFromZero))
Dim bFrac = CInt(Math.Round(bLen / (aLen + bLen) * totalLength, MidpointRounding.AwayFromZero))
' ensure there is at least one character from each string...
If aFrac = 0 Then
aFrac = 1
bFrac -= 1
End If
If bFrac = 0 Then
bFrac = 1
aFrac -= 1
End If
Dim aPart = a.Substring(0, aFrac)
Dim bPart = b.Substring(0, bFrac)
Return aPart & bPart
End Function
Sub Main()
Dim a = New String("A"c, 10)
Dim b = New String("b"c, 40)
Dim c = CombineWithLengthConstraint(a, b, 10)
Console.WriteLine(c)
Console.WriteLine(Len(c))
Console.ReadLine()
End Sub
End Module
Outputs:
AAbbbbbbbb
10
As you can see, the first string, which was 1/5 of the total number of characters, ended up contributing 1/5 of the result.
The VB.NET Len function gives 0 if its argument is Nothing.
I tested it as working with all lengths from 0 to 100 of both strings being combined into one string of length 100 just in case I had made a mistake with the rounding or anything.
Of course, you could return, say, the ending part of string b instead of the starting part if that made sense in the particular application.
Although not exactly what you asked for, here's another option...
Public Function WeirdConcatinate(a As String, b As String) As String
Dim totalLen = a.Length + b.Length
If totalLen > 100 Then
Dim aLen = 100 * a.Length \ totalLen
Dim bLen = 100 - aLen
Return a.Remove(aLen) & b.Remove(bLen)
Else
Return a & b
End If
End Function
This will give you a number of characters from each string (approximately) proportional to how long they are compared to each other. If both strings are the same length, you'll get 50 from each. If a.Length = 100 and b.Length = 50, you'll end up with 66 from a and 34 from b.
Truncate them after concatenating them, then:
Dim c = a & b
If c.Length > 100 Then c = c.Remove(100)
If you want to preserve as much as possible of the start of each string:
Dim c = ""
If(a.Length > 50 AndAlso b.Length < 50)
c = a.Remove(100 - b.Length) & b
Else If a.Length > 50 AndAlso b.Length > 50
c= a.Remove(50) & b.Remove(50)
Else
c = a & b
End if
If c.Length > 100 Then c = c.Remove(100)
As with some other answers, the algorithm is open to interpretation. My method takes from each string until 100 total characters are taken or the string runs out of characters.
Private Function concat(a As String, b As String, length As Integer) As String
Dim ca As New System.Text.StringBuilder()
Dim cb As New System.Text.StringBuilder()
For i As Integer = 0 To length - 1
ca.Append(If(i >= a.Length, "", a(i)))
cb.Append(If(i >= b.Length, "", b(i)))
If ca.Length + cb.Length >= length Then Exit For
Next
Return (ca.ToString() & cb.ToString() & New String(" "c, 100)).Substring(0, length)
End Function
Sub Main()
Dim a As String = New String("a"c, 0)
Dim b As String = New String("b"c, 5)
Dim c As String = concat(a, b, 100)
Console.WriteLine($"'{c}'")
End Sub
'bbbbb '
(padded to 100 characters, doesn't render in block quote)
Dim a As String = New String("a"c, 30)
Dim b As String = New String("b"c, 90)
'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb'
Dim a As String = New String("a"c, 72)
Dim b As String = New String("b"c, 64)
'aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb'
(your example in a comment. 72 >> 50, 64 >> 50)
I have a string S which consists of a's and b's. Perform the below operation once. Objective is to obtain the lexicographically smallest string.
Operation: Reverse exactly one substring of S
e.g.
if S = abab then Output = aabb (reverse ba of string S)
if S = abba then Output = aabb (reverse bba of string S)
My approach
Case 1: If all characters of the input string are same then output will be the string itself.
Case 2: if S is of the form aaaaaaa....bbbbbb.... then answer will be S itself.
otherwise: Find the first occurence of b in S say the position is i. String S will look like
aa...bbb...aaaa...bbbb....aaaa....bbbb....aaaaa...
|
i
In order to obtain the lexicographically smallest string the substring that will be reversed starts from index i. See below for possible ending j.
aa...bbb...aaaa...bbbb....aaaa....bbbb....aaaaa...
| | | |
i j j j
Reverse substring S[i:j] for every j and find the smallest string.
The complexity of the algorithm will be O(|S|*|S|) where |S| is the length of the string.
Is there a better way to solve this problem? Probably O(|S|) solution.
What I am thinking if we can pick the correct j in linear time then we are done. We will pick that j where number of a's is maximum. If there is one maximum then we solved the problem but what if it's not the case? I have tried a lot. Please help.
So, I came up with an algorithm, that seems to be more efficient that O(|S|^2), but I'm not quite sure of it's complexity. Here's a rough outline:
Strip of the leading a's, storing in variable start.
Group the rest of the string into letter chunks.
Find the indices of the groups with the longest sequences of a's.
If only one index remains, proceed to 10.
Filter these indices so that the length of the [first] group of b's after reversal is at a minimum.
If only one index remains, proceed to 10.
Filter these indices so that the length of the [first] group of a's (not including the leading a's) after reversal is at a minimum.
If only one index remains, proceed to 10.
Go back to 5, except inspect the [second/third/...] groups of a's and b's this time.
Return start, plus the reversed groups up to index, plus the remaining groups.
Since any substring that is being reversed begins with a b and ends in an a, no two hypothesized reversals are palindromes and thus two reversals will not result in the same output, guaranteeing that there is a unique optimal solution and that the algorithm will terminate.
My intuition says this approach of probably O(log(|S|)*|S|), but I'm not too sure. An example implementation (not a very good one albeit) in Python is provided below.
from itertools import groupby
def get_next_bs(i, groups, off):
d = 1 + 2*off
before_bs = len(groups[i-d]) if i >= d else 0
after_bs = len(groups[i+d]) if i <= d and len(groups) > i + d else 0
return before_bs + after_bs
def get_next_as(i, groups, off):
d = 2*(off + 1)
return len(groups[d+1]) if i < d else len(groups[i-d])
def maximal_reversal(s):
# example input: 'aabaababbaababbaabbbaa'
first_b = s.find('b')
start, rest = s[:first_b], s[first_b:]
# 'aa', 'baababbaababbaabbbaa'
groups = [''.join(g) for _, g in groupby(rest)]
# ['b', 'aa', 'b', 'a', 'bb', 'aa', 'b', 'a', 'bb', 'aa', 'bbb', 'aa']
try:
max_length = max(len(g) for g in groups if g[0] == 'a')
except ValueError:
return s # no a's after the start, no reversal needed
indices = [i for i, g in enumerate(groups) if g[0] == 'a' and len(g) == max_length]
# [1, 5, 9, 11]
off = 0
while len(indices) > 1:
min_bs = min(get_next_bs(i, groups, off) for i in indices)
indices = [i for i in indices if get_next_bs(i, groups, off) == min_bs]
# off 0: [1, 5, 9], off 1: [5, 9], off 2: [9]
if len(indices) == 1:
break
max_as = max(get_next_as(i, groups, off) for i in indices)
indices = [i for i in indices if get_next_as(i, groups, off) == max_as]
# off 0: [1, 5, 9], off 1: [5, 9]
off += 1
i = indices[0]
groups[:i+1] = groups[:i+1][::-1]
return start + ''.join(groups)
# 'aaaabbabaabbabaabbbbaa'
TL;DR: Here's an algorithm that only iterates over the string once (with O(|S|)-ish complexity for limited string lengths). The example with which I explain it below is a bit long-winded, but the algorithm is really quite simple:
Iterate over the string, and update its value interpreted as a reverse (lsb-to-msb) binary number.
If you find the last zero of a sequence of zeros that is longer than the current maximum, store the current position, and the current reverse value. From then on, also update this value, interpreting the rest of the string as a forward (msb-to-lsb) binary number.
If you find the last zero of a sequence of zeros that is as long as the current maximum, compare the current reverse value with the current value of the stored end-point; if it is smaller, replace the end-point with the current position.
So you're basically comparing the value of the string if it were reversed up to the current point, with the value of the string if it were only reversed up to a (so-far) optimal point, and updating this optimal point on-the-fly.
Here's a quick code example; it could undoubtedly be coded more elegantly:
function reverseSubsequence(str) {
var reverse = 0, max = 0, first, last, value, len = 0, unit = 1;
for (var pos = 0; pos < str.length; pos++) {
var digit = str.charCodeAt(pos) - 97; // read next digit
if (digit == 0) {
if (first == undefined) continue; // skip leading zeros
if (++len > max || len == max && reverse < value) { // better endpoint found
max = len;
last = pos;
value = reverse;
}
} else {
if (first == undefined) first = pos; // end of leading zeros
len = 0;
}
reverse += unit * digit; // update reverse value
unit <<= 1;
value = value * 2 + digit; // update endpoint value
}
return {from: first || 0, to: last || 0};
}
var result = reverseSubsequence("aaabbaabaaabbabaaabaaab");
document.write(result.from + "→" + result.to);
(The code could be simplified by comparing reverse and value whenever a zero is found, and not just when the end of a maximally long sequence of zeros is encountered.)
You can create an algorithm that only iterates over the input once, and can process an incoming stream of unknown length, by keeping track of two values: the value of the whole string interpreted as a reverse (lsb-to-msb) binary number, and the value of the string with one part reversed. Whenever the reverse value goes below the value of the stored best end-point, a better end-point has been found.
Consider this string as an example:
aaabbaabaaabbabaaabaaab
or, written with zeros and ones for simplicity:
00011001000110100010001
We iterate over the leading zeros until we find the first one:
0001
^
This is the start of the sequence we'll want to reverse. We will start interpreting the stream of zeros and ones as a reversed (lsb-to-msb) binary number and update this number after every step:
reverse = 1, unit = 1
Then at every step, we double the unit and update the reverse number:
0001 reverse = 1
00011 unit = 2; reverse = 1 + 1 * 2 = 3
000110 unit = 4; reverse = 3 + 0 * 4 = 3
0001100 unit = 8; reverse = 3 + 0 * 8 = 3
At this point we find a one, and the sequence of zeros comes to an end. It contains 2 zeros, which is currently the maximum, so we store the current position as a possible end-point, and also store the current reverse value:
endpoint = {position = 6, value = 3}
Then we go on iterating over the string, but at every step, we update the value of the possible endpoint, but now as a normal (msb-to-lsb) binary number:
00011001 unit = 16; reverse = 3 + 1 * 16 = 19
endpoint.value *= 2 + 1 = 7
000110010 unit = 32; reverse = 19 + 0 * 32 = 19
endpoint.value *= 2 + 0 = 14
0001100100 unit = 64; reverse = 19 + 0 * 64 = 19
endpoint.value *= 2 + 0 = 28
00011001000 unit = 128; reverse = 19 + 0 * 128 = 19
endpoint.value *= 2 + 0 = 56
At this point we find that we have a sequence of 3 zeros, which is longer that the current maximum of 2, so we throw away the end-point we had so far and replace it with the current position and reverse value:
endpoint = {position = 10, value = 19}
And then we go on iterating over the string:
000110010001 unit = 256; reverse = 19 + 1 * 256 = 275
endpoint.value *= 2 + 1 = 39
0001100100011 unit = 512; reverse = 275 + 1 * 512 = 778
endpoint.value *= 2 + 1 = 79
00011001000110 unit = 1024; reverse = 778 + 0 * 1024 = 778
endpoint.value *= 2 + 0 = 158
000110010001101 unit = 2048; reverse = 778 + 1 * 2048 = 2826
endpoint.value *= 2 + 1 = 317
0001100100011010 unit = 4096; reverse = 2826 + 0 * 4096 = 2826
endpoint.value *= 2 + 0 = 634
00011001000110100 unit = 8192; reverse = 2826 + 0 * 8192 = 2826
endpoint.value *= 2 + 0 = 1268
000110010001101000 unit = 16384; reverse = 2826 + 0 * 16384 = 2826
endpoint.value *= 2 + 0 = 2536
Here we find that we have another sequence with 3 zeros, so we compare the current reverse value with the end-point's value, and find that the stored endpoint has a lower value:
endpoint.value = 2536 < reverse = 2826
so we keep the end-point set to position 10 and we go on iterating over the string:
0001100100011010001 unit = 32768; reverse = 2826 + 1 * 32768 = 35594
endpoint.value *= 2 + 1 = 5073
00011001000110100010 unit = 65536; reverse = 35594 + 0 * 65536 = 35594
endpoint.value *= 2 + 0 = 10146
000110010001101000100 unit = 131072; reverse = 35594 + 0 * 131072 = 35594
endpoint.value *= 2 + 0 = 20292
0001100100011010001000 unit = 262144; reverse = 35594 + 0 * 262144 = 35594
endpoint.value *= 2 + 0 = 40584
And we find another sequence of 3 zeros, so we compare this position to the stored end-point:
endpoint.value = 40584 > reverse = 35594
and we find it has a smaller value, so we replace the possible end-point with the current position:
endpoint = {position = 21, value = 35594}
And then we iterate over the final digit:
00011001000110100010001 unit = 524288; reverse = 35594 + 1 * 524288 = 559882
endpoint.value *= 2 + 1 = 71189
So at the end we find that position 21 gives us the lowest value, so it is the optimal solution:
00011001000110100010001 -> 00000010001011000100111
^ ^
start = 3 end = 21
Here's a C++ version that uses a vector of bool instead of integers. It can parse strings longer than 64 characters, but the complexity is probably quadratic.
#include <vector>
struct range {unsigned int first; unsigned int last;};
range lexiLeastRev(std::string const &str) {
unsigned int len = str.length(), first = 0, last = 0, run = 0, max_run = 0;
std::vector<bool> forward(0), reverse(0);
bool leading_zeros = true;
for (unsigned int pos = 0; pos < len; pos++) {
bool digit = str[pos] - 'a';
if (!digit) {
if (leading_zeros) continue;
if (++run > max_run || run == max_run && reverse < forward) {
max_run = run;
last = pos;
forward = reverse;
}
}
else {
if (leading_zeros) {
leading_zeros = false;
first = pos;
}
run = 0;
}
forward.push_back(digit);
reverse.insert(reverse.begin(), digit);
}
return range {first, last};
}
I am trying to figure out a way to make a for loop in which I can compare two cells that will give me two different means. One for class char and the other for class double.
This is what I have so far.
V = {2; 'tree'; 3; 'hope'};
W = {2; 'tree'; 3; 'hope'};
for i = 1:length(V);
if isequal(class(V{i}), 'double')
num = V{i}
elseif isequal(class(V{i}), 'char')
str = V{i}
end
end
for i = 1:length(W);
if isequal(class(W{i}), 'double')
acc_n(i) = isequal(V{i}, W{i})
elseif isequal(class(W{i}), 'char')
acc_s(i) = strcmp(V{i}, W{i})
end
end
mean_d = mean(acc_n)
mean_s = mean(acc_s)
The output I get is:
acc_n =
1 0 1
acc_s =
0 1 0 1
mean_d =
0.6667
mean_s =
0.5000
The output I want is:
1 1 for string, mean = 1. 1 1 for double, mean = 1
How can I do a loop where it only takes the numbers of the cell and the words of the cell separately?
Is there any possible way to only loop through the words or the numbers?
You can first extract strings and doubles and treat them separately, that will avoid loops.
V = {2; 'tree'; 3; 'hope'};
W = {2; 'tree'; 3; 'hope'};
VChar=V(cellfun(#ischar,V));
WChar=W(cellfun(#ischar,W));
acc_s=VChar==WChar;
VNum=cell2mat(V(cellfun(#isnumeric,V)));
WNum=cell2mat(W(cellfun(#isnumeric,W)));
acc_n=VNum==WNum;
Loop version: I haven't tested this but it should work.
%Assumes V and W have equal number of elements.
acc_n=[];
acc_s=[];
for i=1:numel(V)
if isequal(class(V{i}), 'double') && isequal(V{i},W{i})
acc_n=[acc_n true];
elseif isequal(class(V{i}), 'char') && strcmp(V{i},W{i})
acc_s=[acc_s true];
end
end