I'm using my gauge-solid max value of 5 but on the chart the max value is 6
Thanks.
Related
Table 1:
Position
Team
1
MCI
2
LIV
3
MAN
4
CHE
5
LEI
6
AST
7
BOU
8
BRI
9
NEW
10
TOT
Table 2
Position
Team
1
LIV
2
MAN
3
MCI
4
CHE
5
AST
6
LEI
7
BOU
8
TOT
9
BRI
10
NEW
Output I'm looking for is
Position difference = 10 as that is the total of the positional difference. How can I do this in excel/google sheets? So the positional difference is always a positive even if it goes up or down. Think of it as a league table.
Table 2 New (using formula to find positional difference):
Position
Team
Positional Difference
1
LIV
1
2
MAN
1
3
MCI
2
4
CHE
0
5
AST
1
6
LEI
1
7
BOU
0
8
TOT
2
9
BRI
1
10
NEW
1
Try this:
=IFNA(ABS(INDEX(A:B,MATCH(E2,B:B,0),1)-D2),"-")
Assuming that table 1 is at columns A:B:
I have a scenario here.
Cycle Values
1 0.5
5 1.7
6 0.65
7 2.5
8 0.14
In Tableau By calculation I need to get the last value 0.14
I was wondering how modulo works. I know how it works when the bigger number comes first, but not the opposite. I know that 7 % 3 = 1 as 3 goes up to 7 2 times and the remaining is 1. However, when it's 3 % 7. I have used the calculator it shows 3. Is this because 7 goes up to 3 zero times and the remaining is 3? Is that how it works?
Your reasoning is correct. Any time the divisor is larger than the dividend, the result of the modulo operation equals the dividend.
7*x + y = 3, x and y are int, and x >= 0,
what y = ?
yes, y = 3.
3 MOD 7 = 0R3
This is so because 3/7 is >0 but <1
Mod just means you take the remainder after performing the division. When you divide 3 by 7 you get 3= 0*7 + 3 which means that the remainder is 3.
7/3 -> 3 goes into 7 2 times, and 7 - (3x2) = 1, so your modulus is 1
3/7 -> 7 goes into 3 zero times, and 3 - 0 = 3, so here your modulus is 3
I've created an info text file by using opencv_annotation tool, with around 300 image, and some contain multiple ROIs (Region of Interests). The following is an example output of the file, with dots indicating many lines with the same format:
positives\1\105.png 1 9 10 17 14
...
positives\2\003.png 2 14 2 5 7 11 18 8 9
...
positives\3\045.png 3 21 9 7 9 13 10 9 11 7 15 6 7
However, opencv_annotation then crashes, with the error assertion failed (ssize.area() > 0) ..., and only a fraction (~200 out of ~600) of the ROIs in the info text file were placed into the vec file, verified by how opencv_traincascade reports insufficient samples when using the parameter -numPos 500 when attempting to use the vec file.
Why does this occur, and how can I fix it?
The cause of this is that a ROI that was improperly defined while creating the samples.
In my specific case, this specific line was found:
positives\2\005.png 2 9 6 6 7 0 0 0 0
Recall how the format of these files:
[filename] [# of objects] [[x y width height] [... 2nd object] ...]
In my case, the specified region had a width and length of 0 pixels, which causes opencv_createsamples to crash with the very error. Simply removing the object data and decrementing the [# of objects] by one solved this issue, like so:
positives\2\005.png 1 9 6 6 7
Additionally, one should also look out for values that reach out of bounds, such as the following:
positives\2\005.png 2 9 6 6 7 -5 0 3 2
positives\2\005.png 2 9 6 6 7 30 30 200 300
positives\2\005.png 2 9 6 6 7 35 35 3 5
In the first example, no point should be negative. In the second example, our .png file is only 32 x 32 pixels large, so a ROI cannot have a width and height of 200 and 300, respectively. Finally, in the last example, the x and y positions are out of bounds of our image.
I wish to make a formula to sum up the value with 2 criteria, example show as below:-
A B C D E
1 1-Apr 2-Apr 3-Apr 4-Apr
2 aa 1 4 7 10
3 bb 2 5 8 11
4 cc 3 6 9 12
5
6 Criteria 1 bb
7 Range start 2-Apr-16
8 Range End 4-Apr-16
9 Total sum #VALUE!
tried formula
1 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(C7,B1:E1,0)))
* Only return 1 cell value
2 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(">="&C7,B1:E1,0)))
* Showed N/A error
3 SUMIFS(B2:E4,A2:A4,C6,B1:E1,">="&C7,B1:E1,"<="&C8)
* Showed #Value error
Hereby I attached a link of picture for better understanding :
Can anyone help me on the formula?
I figured out the solution with step evaluation:
=SUMIF(B1:F1,">="&C7,INDEX(B2:F4,MATCH(C6,A2:A4,0),0)) -
SUMIF(B1:F1,">"&C8,INDEX(B2:F4,MATCH(C6,A2:A4,0),0))