I've got hiking distance data from a start point in column A and a column with a yes/no condition (let's say a "Y" denotes a campsite, for example).
What I'm trying to achieve is to calculate the distance between each distance marker in column A that has the condition "Y" in column B. (Desired output is column C.)
A B C
--------------
0 Y
12
26 Y 26 (26 - 0 = 26)
57
124 Y 98 (124 - 26 = 98)
137
152 Y 28 (152 - 124 = 28)
169
. . .
. . .
. . .
I can pull out the distance from column A with a simple IF statement, but that doesn't get me anywhere, of course.
I've searched the Internet extensively and there are a ton of threads out there about finding the last value or last non-empty value in a column.
So I've tried to use INDEX, FILTER, and LOOKUP in all sorts of combinations, but sadly nothing produces the result I'm looking for.
The tricky part, I guess, is to find the last value with a Y above the "current" Y (if that makes any sense).
In C2 try
=ArrayFormula(if(B2:B="y", A2:A-iferror(vlookup(row(A2:A)-1, filter({row(A2:A), A2:A}, len(B2:B)),2)),))
and see if that works?
Related
I want to make the total of values every each member or names in every each their own group at the first match (or after blank space) or highest values positions of each them on column "D" according to column "B" with the result's row of an output like the exactly as an EXPECT OUTPUT as act of what I've just created on column "E". That's the replace a little bit down of just only one row from the column "B" positions or row must be the same as the column "C" and "D". Could we do this anyway ?
My achievements: I feel I've tried this before and got succeed to achieve this but I've forgot how to solve this when that happened. But it's look like kinda this code of formula:
=FILTER(IF(IFERROR(MATCH($B$3:$B;$B:$B;0);0)=ROW($B$3:$B);SUMIF($B$3:$B;$B$3:$B;$D$3:$D);"");$B$3:$B<>"0")
I don't know if I'm right or wrong but please see the table I'd created at the down below this and also see how I expected with that and feel free as well to edit to my doc file of google sheet I attached down below this.
THIS HERE YOU CAN EDIT TO MY SAMPLE G.SHEET TO SOLVE THIS MY QUIZ. THANKS IN ADVANCE!
A
B
C
D
E
1
2
N U M B
I D - M E M B E R
I D - C O D E
V A L U E S
E X P E C T O U T P U T
3
4
4
JYFI7
5
JYFI7
J3573
3
6
6
JYFI7
IYR
1
7
JYFI7
F498S
2
8
9
3
DFJ9F11
10
DFJ9F11
C684J
7
8
11
DFJ9F11
J58
1
12
13
2
H684K
14
H684K
JF585
2
2
15
16
1
FJSR
17
FJSR
4684
7
16
18
FJSR
834
1
19
FJSR
49
2
20
FJSR
9835
6
Here's a possible solution:
=ARRAYFORMULA(LAMBDA(cusum,IF(SCAN(,cusum,
LAMBDA(acc,cur,if(cur="",,acc+1)))=1,cusum,))
(SORT(SCAN(,SORT(D3:D,ROW(D3:D),0),
LAMBDA(acc,cur,if(cur="",,acc+cur))),ROW(D3:D),0)))
You can find it in tab 'z' cell F3.
I have a constant number X. I also have two numbers that add up to it. How can I make it so that if I change one number, the other number automatically changes so that it still adds up to X.
I have tried to take subtract the one number from X and add it to the other number, but instead I got two numbers in the thousands.
Assuming your constant value is 10, you can set this in a cell and make all your other calculations based on it.
For example, you can have cell C2 containing your constant, in this example, 10
Then in C4 you can have the number which you change, and the value of C5 will be equal to the value of the constant minus the value in C4.
You can then finally do your sum wherever you want, adding up the values of C4 and C5.
Here's an example Spreadsheet:
Untitiled spreadsheet ☆
File Edit View Insert Format Data Tools Extensions Help Last edit was 2 minutes ago
↶ ↷ 🖶 ⮷ | 100%⯆ | $ % .0 .00 123⯆ | Default(Ro... ⯆ | 10 ⯆ | B | I | S | A |⯐|☰
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
1
2
Contsant:
10
3
4
Number 1:
3
5
Number 2:
=(C2 - C4)
6
7
Sum:
=(C4 + C5)
8
Minimum number states in the DFA accepting strings (base 3 i.e,, ternary form) congruent to 5 modulo 6?
I have tried but couldn't do it.
At first sight, It seems to have 6 states but then it can be minimised further.
Let's first see the state transition table:
Here, the states q0, q1, q2,...., q5 corresponds to the states with modulo 0,1,2,..., 5 respectively when divided by 6. q0 is our initial state and since we need modulo 5 therefore our final state will be q5
Few observations drawn from above state transition table:
states q0, q2 and q4 are exactly same
states q1, q3 and q5 are exactly same
The states which make transitions to the same states on the same inputs can be merged into a single state.
Note: Final and Non-final states can never be merged.
Therefore, we can merge q0, q2, q4 together and q1, q3 together leaving the state q5 aloof from collation.
The final Minimal DFA has 3 states as shown below:
Let's look at a few strings in the language:
12 = 1*3 + 2 = 5 ~ 5 (mod 6)
102 = 1*9 + 0*3 + 2 = 11 ~ 5 (mod 6)
122 = 1*9 + 2*3 + 2 = 17 ~ 5 (mod 6)
212 = 2*9 + 1*3 + 2 = 23 ~ 5 (mod 6)
1002 = 1*18 + 0*9 + 0*9 + 2 = 29 ~ 5 (mod 6)
We notice that all the strings end in 2. This makes sense since 6 is a multiple of 3 and the only way to get 5 from a multiple of 3 is to add 2. Based on this, we can try to solve the problem of strings congruent to 3 modulo 6:
10 = 3
100 = 9
120 = 15
210 = 21
1000 = 27
There's not a real pattern emerging, but consider this: every base-3 number ending in 0 is definitely divisible by 3. The ones that are even are also divisible by 6; so the odd numbers whose base-3 representation ends in 0 must be congruent to 3 mod 6. Because all the powers of 3 are odd, we know we have an odd number if the number of 1s in the string is odd.
So, our conditions are:
the string begins with a 1;
the string has an odd number of 1s;
the string ends with 2;
the string can contain any number of 2s and 0s.
To get the minimum number of states in such a DFA, we can use the Myhill-Nerode theorem beginning with the empty string:
the empty string can be followed by any string in the language. Call its equivalence class [e]
the string 0 cannot be followed by anything since valid base-3 representations don't have leading 0s. Call its equivalence class [0].
the string 1 must be followed with stuff that has an even number of 1s in it ending with a 2. Call its equivalence class [1].
the string 2 can be followed by anything in the language. Indeed, you can verify that putting a 2 at the front of any string in the language gives another string in the language. However, it can also be followed by strings beginning with 0. Therefore, its class is new: [2].
the string 00 can't be followed by anything to fix it; its class is the same as its prefix 0, [0]. same for the string 01.
the string 10 can be followed by any string with an even number of 1s that ends in a 2; it is therefore equivalent to the class [1].
the string 11 can be followed by any string in the language whatever; indeed, you can verify prepending 11 in front of any string in the language gives another solution. However, it can also be followed by strings beginning with 0. Therefore, its class is the same as [2].
12 can be followed by a string with an even number of 1s ending in 2, as well as by the empty string (since 12 is in fact in the language). This is a new class, [12].
21 is equivalent to 1; class [1]
22 is equivalent to 2; class [2]
20 is equivalent to 2; class [2]
120 is indistinguishable from 1; its class is [1].
121 is indistinguishable from [2].
122 is indistinguishable from [12].
We have seen no new equivalence classes on new strings of length 3; so, we know we have seen all the equivalence classes. They are the following:
[e]: any string in the language can follow this
[0]: no string can follow this
[1]: a string with an even number of 1s ending in 2 can follow this
[2]: same as [e] but also strings beginning with 0
[12]: same as [1] but also the empty string
This means that a minimal DFA for our language has five states. Here is the DFA:
[0]
^
|
0
|
----->[e]--2-->[2]<-\
| ^ |
| | |
1 __1__/ /
| / /
| | 1
V V |
[1]--2-->[12]
^ |
| |
\___0___/
(transitions not pictured are self-loops on the respective states).
Note: I expected this DFA to have 6 states, as Welbog pointed out in the other answer, so I might have missed an equivalence class. However, the DFA seems right after checking a few examples and thinking about what it's doing: you can only get to accepting state [12] by seeing a 2 as the last symbol (definitely necessary) and you can only get to state [12] from state [1] and you must have seen an odd number of 1s to get to [1]…
The minimum number of states for almost all modulus problems is the base of the modulus. The general strategy is one state for every modulus, as transitions between moduli are independent of what the previous numbers were. For example, if you're in state r4 (representing x = 4 (mod 6)), and you encounter a 1 as your next input, your new modulus is 4x6+1 = 25 = 1 (mod 6), so the transition from r4 on input 1 is to r1. You'll find that the start state and r0 can be merged, for a total of 6 states.
As im so new to this field and im trying to explore the data for a time series, and find the missing values and count them and study a distribution of their length and fill in these gaps, the thing is i have, let's say 10 file.txt and for each file i have 2 columns as follows:
C1 C2
944 0
920 1
920 2
928 3
912 7
920 8
920 9
880 10
888 11
920 12
944 13
and so on... lets say till 100 and not necessarily the 10 files have the same number of observations.
so here for example the missing values and not necessarily appears in all files that i have, missing value are: 4,5 and 6 in C2 and the corresponding 1st column C1(measured in milliseconds, so the value of 928ms is not a time neighbor of 912ms). So i want to find those gaps(the total missing values in all 10 files) and show a histogram of their lengths.
i wrote a piece of code in R, but the problem is that i don't get the exact total number that i should have for the missing values.
path = "files path"
out.file<-data.frame(TS = 0, Index = 0, File = '')
file.names <- dir(path, pattern =".txt")
for(i in 1:length(file.names)){
file <- cbind(read.table(file.names[i],
header=F,
sep ="\t",
stringsAsFactors=FALSE),
file.names[i])
colnames(file) <- c('TS', 'Index', 'File')
out.file <- rbind(out.file, file)
}
d = dim(out.file)[1]
misDa = 0
for(i in 2:(d-1)){
if(abs(out.file$Index[i]-out.file$Index[i+1]) > 1)
misDa = misDa+1
}
Hard to give specific hints without having a more extensive example of your data that contains some of the actual NAs.
If you are using R (like it seems) the naniar and the imputeTS packages offer nice functions for missing data visualizations.
Some examples from the naniar package, which is especially good for multivariate data (more plot examples):
Some examples from the imputeTS package, which is especially good for time series data (additional plot examples):
Example
3 2 5 5
a b c d
Joining first two
5 | 5 5
3 2 | c d
a b |
I have to put the new tree of five into the queue
Am I obligated to put it in the end like this:
5 5 5
c d / \
3 2
a b
Or can I put it in the beginning:
5 5 5
3 2 c d
a b
Or even in the middle of 'c' and 'd'
Is it my choice or is there a rule?
It's not your choice, the Queue needs to be sorted at all times (by it's number of occurrences and in case of equal number of occurrences by the depth of the tree). So it needs to be inserted where it belongs into the order.
This is needed to pick the sub-trees with the least amount of occurrences and if there is choice the most shallow one of them by simply pop-ing them.
If you simply resort after every insertion (this is inefficient and should not be done) the position obviously doesn't matter.
Yes, it's your choice. Whichever way you will get an optimal Huffman code, even though two resulting codes can be manifestly different.
You can get:
a - 00
b - 01
c - 10
d - 11
or you can get:
a - 111
b - 110
c - 10
d - 0
Now if I multiply the number of bits in each symbol times the number of occurrences, I get for the first code: 2*3 + 2*2 + 2*5 + 2*5 = 30 bits. For the second code: 3*3 + 3*2 + 2*5 + 1*5 = 30 bits. So both codes will code the original message to exactly 30 bits.