I have this LibreOffice calc file with raws with full of zero
raw1 raw2 raw3 raw4 raw5 raw6 raw7 raw8 raw9
0 0 0 0 C 0 0 0 0
0 0 0 0 0 0 0 W 0
I want to print only the character inside the row, like this
Result
C
W
I did try with 'if' condition
IF(CD2:CR16 = 1, CD2:CR16)
but it's give me an error
Use MATCH to find the column that contains a character, and then INDEX to get the cell's value.
=INDEX(CD2:CR2, MATCH("[A-Z]", CD2:CR2, 0))
For this to work, go to Tools -> Options -> LibreOffice Calc -> Calculate, and choose Enable regular expressions in formulas.
EDIT:
According to https://help.libreoffice.org/Common/List_of_Regular_Expressions, [:print:] represents any printable character, so it grabs the first zero, which is probably why it does not seem to do what you want.
To match one of several words, the regular expression should be like this:
"word1|word2|word3"
Or for any word consisting of one or more letters:
"[:alpha:]+"
EDIT 2:
To grab C and 8 from 0 0 C 0 and 8 0 0 0 respectively, use "[A-Z1-9]".
Related
I have a spreadsheet and in one of the tabs I have a table with computed data from other tabs. This is small table with 11 columns. Row(1) is the Header row and Column A is the list of items, Column B to J is the types. Data consists of numbers only.
As the data is computed, time to time values in some of the columns thru B to J can be totally zero. I want to create a subset of this table with QUERY but constructing a dynamic range getting only the columns which has at least 1 value which is greater than zero.
I'm aware that a range can be created as an array like {A:A\B:B\D:D} but in my case I don't know which columns can have values of greater than zero and I don't want to take columns into the range which has completely zero values.
I have created an expression to concatenate this array value as a text in a cell, however I can't use it with the QUERY formula either with INDEX or TEXT functions. Table is like this:
Items TypeA TypeB TypeC TypeD
Bronze 0 0 0 0
Silver 0 0 1 0
Gold 0 0 1 0
Titanimum 1 0 0 0
For this snapshot of table, I want to QUERY range to be {A:A\B:B\D:D}. However, as the data is computed, the table can be like this after 2hrs or the next day:
Items TypeA TypeB TypeC TypeD
Bronze 1 0 0 1
Silver 0 0 1 0
Gold 0 1 1 0
Titanimum 1 0 0 0
And so, for this snapshot of table, I want to QUERY range to be {A:A\B:B\C:C\D:D\E:E}.
Is this doable? And how can I achieve or construct a dynamic QUERY range?
Thanks for everyone...
You can remove columns from a range based on a criteria using the FILTER command.
Unfiltered
Items TypeA TypeB TypeC TypeD TypeE TypeF TypeG
Bronze 1 0 0 1 0 0 1
Silver 1 1 0 1 0 0 1
Gold 1 0 0 1 0 0 1
Titan 1 0 0 1 1 0 1
1 4 1 0 4 1 0 4
Filtered to remove columns with total of 0
Items TypeA TypeB TypeD TypeE TypeG
Bronze 1 0 1 0 1
Silver 1 1 1 0 1
Gold 1 0 1 0 1
Titan 1 0 1 1 1
The 'trick' is to sum the sum the column data (for your example) and then test for >0
The filter expression is:
=FILTER(A1:H5,A6:H6 >0)
By way of explanation:
A1:H5 is the range to be filtered;
A6:H6 >0 selects all columns that have a value > 0 in row 6
I placed a 1 in A6 to make sure colA is included.
You can now do queries on the range returned by the above expression.
There are two MxN 2D arrays:
rand bit [M-1:0] src [N-1:0];
rand bit [M-1:0] dst [N-1:0];
Both of them will be randomized separately so that they both have P number of 1'b1 in them and rest are 1'b0.
A third MxN array of integers named 'map' establishes a one to one mapping between the two arrays 'src' and 'dst'.
rand int [M-1:0] map [N-1:0];
Need a constraint for 'map' such that after randomization, for each element of src[i][j] where src[i][j] == 1'b1, map[i][j] == M*k+l when dst[k][l] == 1. The k and l must be unique for each non-zero element of map.
To give an example:
Let M = 3 and N = 2.
Let src be
[1 0 1
0 1 0]
Let dst be
[0 1 1
1 0 0]
Then one possible randomization of 'map' will be:
[3 0 1
0 2 0]
In the above map:
3 indicates pointing from src[0,0] to dst[1,0] (3 = 1*M+0)
1 indicates pointing from src[0,2] to dst[0,1] (1 = 0*M+1)
2 indicates pointing from src[1,1] to dst[0,2] (2 = 0*M+2)
This is very difficult to express as a SystemVerilog constraint because
there is no way to conditionally select elements of an array to be unique
You cannot have random variables as part of index expression to an array element.
Since you are randomizing src and dst separately, it might be easier to compute the pointers and then randomly choose the pointers to fill in the map.
module top;
parameter M=3,N=4,P=4;
bit [M-1:0] src [N];
bit [M-1:0] dst [N];
int map [N][M];
int pointers[$];
initial begin
assert( randomize(src) with {src.sum() with ($countones(item)) == P;} );
assert( randomize(dst) with {dst.sum() with ($countones(item)) == P;} );
foreach(dst[K,L]) if (dst[K][L]) pointers.push_back(K*M+L);
pointers.shuffle();
foreach(map[I,J]) map[I][J] = pointers.pop_back();
$displayb("%p\n%p",src,dst);
$display("%p",map);
end
endmodule
I'm looking at the example: https://github.com/fchollet/keras/blob/master/examples/conv_lstm.py
This RNN is actually predicting the next frame of the movie, so the output should be a movie too (according to the test data fed in). I wonder if there are information lost due to the conv layers with padding.
For example, the underlying Tensorflow is padding bottom right, if there is a big padding: (n stands for numbers)
n n n n 0 0 0
n n n n 0 0 0
n n n n 0 0 0
n n n n 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
when we do the second conv, the bottom right corner will always be 0, which means the back propagation will never be able to capture anything there. As in this case a movie(a square moves on the whole screen), will it lost the information when the validation label is on the bottom right corner?
The answer is yes after asking a Ph.D. doing AI research.
I came across these two lines (back-to-back) of code in a torch project:
im4[{1,{},{}}] = im3[{3,{},{}}]
im4[{3,{},{}}] = im3[{1,{},{}}]
What do these two lines do? I assumed they did some sort of swapping.
This is covered in indexing in the Torch Tensor Documentation
Indexing using the empty table {} is shorthand for all indices in that dimension. Below is a demo which uses {} to copy an entire row from one matrix to another:
> a = torch.Tensor(3, 3):fill(0)
0 0 0
0 0 0
0 0 0
> b = torch.Tensor(3, 3)
> for i=1,3 do for j=1,3 do b[i][j] = (i - 1) * 3 + j end end
> b
1 2 3
4 5 6
7 8 9
> a[{1, {}}] = b[{3, {}}]
> a
7 8 9
0 0 0
0 0 0
This assignment is equivalent to: a[1] = b[3].
Your example is similar:
im4[{1,{},{}}] = im3[{3,{},{}}]
im4[{3,{},{}}] = im3[{1,{},{}}]
which is more clearly stated as:
im4[1] = im3[3]
im4[3] = im3[1]
The first line assigns the values from im3's third row (a 2D sub-matrix) to im4's first row and the second line assigns the first row of im3 to the third row of im4.
Note that this is not a swap, as im3 is never written and im4 is never read from.
This seemed like a trivial question to me, but I cannot get it done correctly. Part of my dataset looks like this
1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0
and contains two “runs” of 1 (not sure if that’s the correct word), one with a length 3, the other with a length of 5.
How can I use Google Docs or similar spreadsheet applications to find the longest of those runs?
In Excel you can use a single formula to get the maximum number of consecutive 1s, i.e.
=MAX(FREQUENCY(IF(A2:A100=1,ROW(A2:A100)),IF(A2:A100<>1,ROW(A2:A100))))
confirmed with CTRL+SHIFT+ENTER
In Google Sheets you can use the same formula but wrap in arrayformula rather than use CSE, i.e.
=arrayformula(MAX(FREQUENCY(IF(A2:A100=1,ROW(A2:A100)),IF(A2:A100<>1,ROW(A2:A100)))))
Assumes data in A2:A100 without blanks
EDIT: whuber's suggestion is just too simple for me to not update this response. One can just use a simple IF statement checking if the current row is equal to 1. If it is, it starts a counter (the prior row + 1), if it is not it starts the counter again at 0.
You just need to initialize the first row of B1 to 1 or 0. Using the dynamic updating of cell formulas once you have it written once it fills in the rest.
So you would start out;
A B
1 1
1 =IF(A2=1, B1+1, 0)
1
0
0
1
1
1
1
0
0
0
Then fill in;
A B
1 1
1 =IF(A2=1, B1+1, 0)
1 =IF(A3=1, B2+1, 0)
0 =IF(A4=1, B3+1, 0)
0 =IF(A5=1, B4+1, 0)
1 =IF(A6=1, B5+1, 0)
1 =IF(A7=1, B6+1, 0)
1 =IF(A8=1, B7+1, 0)
1 =IF(A9=1, B8+1, 0)
0 =IF(A10=1, B9+1, 0)
0 =IF(A11=1, B10+1, 0)
0 =IF(A12=1, B11+1, 0)
And here the result in column B is;
A B
1 1
1 2
1 3
0 0
0 0
1 1
1 2
1 3
1 4
0 0
0 0
0 0
Hopefully the logic is extendable to Google Docs.