I am trying to understand the the MCTAL output of a spherical TMESH tally. What I want is to create one tally bin that has the following boundaries 1.9 cm and 2.1 cm in the radial direction, 88 to 92 degrees in theta and 180 to 360 degrees in the phi direction. my input for the tally is
C tally card spherical mesh energy tally
TMESH
SMESH1:p DOSE 1 1 1 1.0 PEDEP MFACT 1 1 0 1.0
CORA1 1.9 2.1
CORB1 88 92
CORC1 180 360
Now what I expect is one result for that volume what I get are eight values as shown below.
ntal 1
1
tally 1 -1 -3
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
f 4 0 1 2 2
1.90000E+00 2.10000E+00
0.00000E+00 8.80000E+01 9.20000E+01
0.00000E+00 1.80000E+02 3.60000E+02
d 1
u 1
s 2
m 1
c 1
e 1
t 1
vals
5.57481E-04 0.0067 7.68088E-09 0.0493 8.24471E-03 0.0046 1.38395E-07 0.0639
5.53931E-04 0.0046 7.44313E-09 0.0287 8.24244E-03 0.0042 1.27868E-07 0.0553
I am assuming that these eight vals correspond to the eight points that that are listed under f. Does TMESH only give one values for individual points on a grid or can it be used to create a volume within which to obtain a result? lastly to what points do what vals correspond to ?
The matrix bellow the vals is true value of your meshtally result.
but
you must load data to Matlab and reshape it to your mesh tally matrix
With your SMESH setup you score both dose and energy deposition. This causes two bins along the segment axis (the "s 2" record in your mctal). Then, you have only 1 bin along the radial direction (1.9-2.1 cm) and actually TWO bins along each of the angular directions (0-88, 88-92, and 0-180, 180-360) which sums up to 2^3 = 8 bins. The mctal file format is described in the manual: it'a 11-dimension loop. In your case only the s, j and k axes are divided, so it's actully a 3D loop (in this exact order: s being the outer, k - the inner loop). Therefore the value for your volume is either the 4th (1.38395E-07 0.0639) or last (1.27868E-07 0.0553) record depending on whether you need dose or energy deposition.
Related
I have a dataset like this,
sample = {'Theme': ['never give a ten','interaction speed','no feedback,premium'],
'cat1': [0,0,0],
'cat2': [0,0,0],
'cat3': [0,0,0],
'cat4': [0,0,0]
}
pd.DataFrame(sample,columns = ['Theme','cat1','cat2','cat3','cat4'])
Theme cat1 cat2 cat3 cat4
0 never give a ten 0 0 0 0
1 interaction speed 0 0 0 0
2 no feedback,premium 0 0 0 0
Now, I need to replace the values in cat columns based on value in Theme. If the Theme column has 'never give a ten', then change cat1 as 1, similarly if the theme column has 'interaction speed', then change cat2 as 1, if the theme column has 'no feedback' in it, change 'cat3' as 1 and for 'premium' change cat4 as 1.
In this sample I have provided 4 categories, I have in total 21 categories. I can do if word in string 21 times for 21 categories, but I am looking for an efficient way to write this in a function, loop every row and go through the logic and update the corresponding columns, can anyone help please?
Thanks in advance.
Here is possible set columns names by categories with Series.str.get_dummies - columns names are sorted:
df1 = df['Theme'].str.get_dummies(',')
print (df1)
interaction speed never give a ten no feedback premium
0 0 1 0 0
1 1 0 0 0
2 0 0 1 1
If need first column in output add DataFrame.join:
df11 = df[['Theme']].join(df['Theme'].str.get_dummies(','))
print (df11)
Theme interaction speed never give a ten no feedback \
0 never give a ten 0 1 0
1 interaction speed 1 0 0
2 no feedback,premium 0 0 1
premium
0 0
1 0
2 1
If order of columns is important add DataFrame.reindex:
#removed posible duplicates with remain ordering
cols = dict.fromkeys([y for x in df['Theme'] for y in x.split(',')]).keys()
df2 = df['Theme'].str.get_dummies(',').reindex(cols, axis=1)
print (df2)
never give a ten interaction speed no feedback premium
0 1 0 0 0
1 0 1 0 0
2 0 0 1 1
cols = dict.fromkeys([y for x in df['Theme'] for y in x.split(',')]).keys()
df2 = df[['Theme']].join(df['Theme'].str.get_dummies(',').reindex(cols, axis=1))
print (df2)
Theme never give a ten interaction speed no feedback \
0 never give a ten 1 0 0
1 interaction speed 0 1 0
2 no feedback,premium 0 0 1
premium
0 0
1 0
2 1
I am trying to create one heatmap using Gnuplot and my data file structure is looked like below:
6 5 4 3 1 0
3 2 2 0 0 1
0 0 0 0 1 0
0 0 0 0 2 3
0 0 1 2 4 3
the cell values are z values and columns represent y-axis and row are x-axes. that means the first value 6 is the z value where the y-axis is 5th position at x label zero. However, while plotting the heat map I am getting a different color which does not correlate with the z value. Also, I am getting five bins for the x-axis (which is supposed to be 6)and 4 bins (which is supposed to be 5) for the y-axis. My simple code is written below:
set pm3d map
splot 'm.txt' matrix
Please help me out of this confused situation.
Thanks.
I have a bag-of-words representation of a corpus stored in an D by W sparse matrix word_freqs. Each row is a document and each column is a word. A given element word_freqs[d,w] represents the number of occurrences of word w in document d.
I'm trying to obtain another D by W matrix not_word_occs where, for each element of word_freqs:
If word_freqs[d,w] is zero, not_word_occs[d,w] should be one.
Otherwise, not_word_occs[d,w] should be zero.
Eventually, this matrix will need to be multiplied with other matrices which might be dense or sparse.
I've tried a number of methods, including:
not_word_occs = (word_freqs == 0).astype(int)
This words for toy examples, but results in a MemoryError for my actual data (which is approx. 18,000x16,000).
I've also tried np.logical_not():
word_occs = sklearn.preprocessing.binarize(word_freqs)
not_word_occs = np.logical_not(word_freqs).astype(int)
This seemed promising, but np.logical_not() does not work on sparse matrices, giving the following error:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all().
Any ideas or guidance would be appreciated.
(By the way, word_freqs is generated by sklearn's preprocessing.CountVectorizer(). If there's a solution that involves converting this to another kind of matrix, I'm certainly open to that.)
The complement of the nonzero positions of a sparse matrix is dense. So if you want to achieve your stated goals with standard numpy arrays you will require quite a bit of RAM. Here's a quick and totally unscientific hack to give you an idea, how many arrays of that sort your computer can handle:
>>> import numpy as np
>>> a = []
>>> for j in range(100):
... print(j)
... a.append(np.ones((16000, 18000), dtype=int))
My laptop chokes at j=1. So unless you have a really good computer even if you can get the complement (you can do
>>> compl = np.ones(S.shape,int)
>>> compl[S.nonzero()] = 0
) memory will be an issue.
One way out may be to not explicitly compute the complement let's call it C = B1 - A, where B1 is the same-shape matrix completely filled with ones and A the adjacency matrix of your original sparse matrix. For example the matrix product XC can be written as XB1 - XA so you have one multiplication with the sparse A and one with B1 which is actually cheap because it boils down to computing row sums. The point here is that you can compute that without computing C first.
A particularly simple example would be multiplication with a one-hot vector. Such a multiplication just selects a column (if multiplying from the right) or row (if multiplying from the left) of the other matrix. Meaning you just need to find that column or row of the sparse matrix and take the complement (for a single slice no problem) and if you do this for a one-hot matrix, as above you needn't compute the complement explicitly.
Make a small sparse matrix:
In [743]: freq = sparse.random(10,10,.1)
In [744]: freq
Out[744]:
<10x10 sparse matrix of type '<class 'numpy.float64'>'
with 10 stored elements in COOrdinate format>
the repr(freq) shows the shape, elements and format.
In [745]: freq==0
/usr/local/lib/python3.5/dist-packages/scipy/sparse/compressed.py:213: SparseEfficiencyWarning: Comparing a sparse matrix with 0 using == is inefficient, try using != instead.
", try using != instead.", SparseEfficiencyWarning)
Out[745]:
<10x10 sparse matrix of type '<class 'numpy.bool_'>'
with 90 stored elements in Compressed Sparse Row format>
If do your first action, I get a warning and new array with 90 (out of 100) nonzero terms. That not is no longer sparse.
In general numpy functions do not work when applied to sparse matrices. To work they have to delegate the task to sparse methods. But even if logical_not worked it wouldn't solve the memory issue.
Here is an example of using Pandas.SparseDataFrame:
In [42]: X = (sparse.rand(10, 10, .1) != 0).astype(np.int64)
In [43]: X = (sparse.rand(10, 10, .1) != 0).astype(np.int64)
In [44]: d1 = pd.SparseDataFrame(X.toarray(), default_fill_value=0, dtype=np.int64)
In [45]: d2 = pd.SparseDataFrame(np.ones((10,10)), default_fill_value=1, dtype=np.int64)
In [46]: d1.memory_usage()
Out[46]:
Index 80
0 16
1 0
2 8
3 16
4 0
5 0
6 16
7 16
8 8
9 0
dtype: int64
In [47]: d2.memory_usage()
Out[47]:
Index 80
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
9 0
dtype: int64
math:
In [48]: d2 - d1
Out[48]:
0 1 2 3 4 5 6 7 8 9
0 1 1 0 0 1 1 0 1 1 1
1 1 1 1 1 1 1 1 1 0 1
2 1 1 1 1 1 1 1 1 1 1
3 1 1 1 1 1 1 1 0 1 1
4 1 1 1 1 1 1 1 1 1 1
5 0 1 1 1 1 1 1 1 1 1
6 1 1 1 1 1 1 1 1 1 1
7 0 1 1 0 1 1 1 0 1 1
8 1 1 1 1 1 1 0 1 1 1
9 1 1 1 1 1 1 1 1 1 1
source sparse matrix:
In [49]: d1
Out[49]:
0 1 2 3 4 5 6 7 8 9
0 0 0 1 1 0 0 1 0 0 0
1 0 0 0 0 0 0 0 0 1 0
2 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 1 0 0
4 0 0 0 0 0 0 0 0 0 0
5 1 0 0 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0
7 1 0 0 1 0 0 0 1 0 0
8 0 0 0 0 0 0 1 0 0 0
9 0 0 0 0 0 0 0 0 0 0
memory usage:
In [50]: (d2 - d1).memory_usage()
Out[50]:
Index 80
0 16
1 0
2 8
3 16
4 0
5 0
6 16
7 16
8 8
9 0
dtype: int64
PS if you can't build the whole SparseDataFrame at once (because of memory constraints), you can use an approach similar to one used in this answer
I am working on a school project, which is a simulation of logical gates. I can implement and run the simulation with ease, but i need help with showing the output.
Right now, i print everything to the console, like this:
sample frequency: 50
###############################################
IN NOT(1) OUT
IN1:0 IN1:3 IN1:5
IN2:0 IN2:0 IN2:0
OUT:3 OUT:5 OUT:0
0 1 -1 -1
50 1 -1 -1
100 1 0 0
150 0 0 0
200 1 1 1
250 1 0 0
300 1 0 0
350 1 0 0 (IN = 1, delay is 1 so we can see
400 0 0 0 the correct output of NOT element in line 400 <-> 350 + 1*50)
450 1 1 1
500 1 0 0
550 1 0 0
600 1 0 0
650 0 0 0
700 0 1 1
750 1 1 1
800 1 0 0
850 1 0 0
900 1 0 0
950 1 0 0
1000 1 0 0
on the left, there is the simulation time (step). In each step, the values are printed out and new set of inputs is generated.
where there is -1, this means undefined output.
The 3rd row ( IN NOT(1) OUT ) means that there are 3 elements, 1 input, 1 NOT gate and an output. The value in brackets means the delay of the element, so an element with delay value of X will show the correct output after X*sample_freq (excluding the 0 time).
The rows after mean:
IN1 - the index of the node that is read as input 1
IN2 - the index of the node that is read as input 2
OUT - the index of the output node
In this situation, the IN is giving its output to node #3. The NOT element reads its input from node #3 and gives some output to node #5. The overall output of this system is the OUT element, which reads from #5.
Here is the file that specifies the topology:
3 (number of elems)
IN 0 0 3 (no inputs for input element obviously)
NOT 3 0 5 (reads from #3 and outputs to #5)
OUT 5 0 0 (reads from #5 and this is the end point of the system)
There can obviously be more elements, IN's and OUT's, but lets stick to this for the sake of simplicity.
And what i want to see as the result is: X-axis tells the simulation time (0 - 1000, step is 50), y axis tells the output value of each element in the system and the elements write their output one above the other, see this picture as an example.
Can you tell me how to create this kind of gnuplot script, that transforms the output of my application into the desired plot?
Thank you!
ok, I have found a solultion myself, here it is:
first, I had to transform the output of the app a bit, so that it looks like this:
0 1 2 4
49 1 2 4
50 1 2 4
99 1 2 4
100 0 2 4
149 0 2 4
150 0 3 5
199 0 3 5
200 1 3 5
249 1 3 5
250 1 2 4
299 1 2 4
300 0 2 5
349 0 2 5
350 1 3 5
399 1 3 5
400 0 2 4
449 0 2 4
450 1 3 5
499 1 3 5
the extra sim time steps make the edges look almost square, I also separated each column by 2 (added 0 to column #2, added 2 to column #3, added 4 to column #4 and so on), so that it is drawn one above each other and the simple command to plot this is:
plot 'out.txt' using 1:2 with lines, 'out.txt' using 1:3 with lines, 'out.txt' using 1:4 with lines
plus some set xtics, set ytics and other cosmetic stuff
now I have to deal with naming the lines with the names of the elements and voila.
I have a string:
sup_pairs = 'BA CE DF EF AE FC GD DA CG EA AB BG'
How can I combine pairs which have the last character of 1 pair is the first character of the follow pairs into strings? And the new strings must contain all of the character 'A','B','C','D','E','F' , 'G', those characters are appeared in the sup_pairs string.
The expected output should be:
S1 = 'BAEFCGD' % because BA will be followed by AE in sup_pairs string, so we combine BAE, and so on...we continue the rule to generate S1
S2 = 'DFCEABG'
If I have AB, BC and BD, the generated strings should be both : ABC and ABD .
If there is any repeated character in the pairs like : AB BC CA CE . We will skip the second A , and we get ABCE .
This, like all good things in life, is a graph problem. Each letter is a node, and each pair is an edge.
First we must transform your string of pairs into a numeric format so we can use the letters as subscripts. I will use A=2, B=3, ..., G=8:
sup_pairs = 'BA CE DF EF AE FC GD DA CG EA AB BG';
p=strsplit(sup_pairs,' ');
m=cell2mat(p(:));
m=m-'?';
A=sparse(m(:,1),m(:,2),1);
The sparse matrix A is now the adjacency matrix (actually, more like an adjacency list) representing our pairs. If you look at the full matrix of A, it looks like this:
>> full(A)
ans =
0 0 0 0 0 0 0 0
0 0 1 0 0 1 0 0
0 1 0 0 0 0 0 1
0 0 0 0 0 1 0 1
0 1 0 0 0 0 1 0
0 1 0 0 0 0 1 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
As you can see, the edge BA, which translates to subscript (3,2) is equal to 1.
Now you can use your favorite implementation of Depth-first Search (DFS) to perform a traversal of the graph from your starting node of choice. Each path from the root to a leaf node represents a valid string. You then transform the path back into your letter sequence:
treepath=[3,2,6,7,4,8,5];
S1=char(treepath+'?');
Output:
S1 = BAEFCGD
Here's a recursive implementation of DFS to get you going. Normally in MATLAB you have to worry about not hitting the default limitation on recursion depth, but you're finding Hamiltonian paths here, which is NP-complete. If you ever get anywhere near the recursion limit, the computation time will be so huge that increasing the depth will be the least of your worries.
function full_paths = dft_all(A, current_path)
% A - adjacency matrix of graph
% current_path - initially just the start node (root)
% full_paths - cell array containing all paths from initial root to a leaf
n = size(A, 1); % number of nodes in graph
full_paths = cell(1,0); % return cell array
unvisited_mask = ones(1, n);
unvisited_mask(current_path) = 0; % mask off already visited nodes (path)
% multiply mask by array of nodes accessible from last node in path
unvisited_nodes = find(A(current_path(end), :) .* unvisited_mask);
% add restriction on length of paths to keep (numel == n)
if isempty(unvisited_nodes) && (numel(current_path) == n)
full_paths = {current_path}; % we've found a leaf node
return;
end
% otherwise, still more nodes to search
for node = unvisited_nodes
new_path = dft_all(A, [current_path node]); % add new node and search
if ~isempty(new_path) % if this produces a new path...
full_paths = {full_paths{1,:}, new_path{1,:}}; % add it to output
end
end
end
This is a normal Depth-first traversal except for the added condition on the length of the path in line 15:
if isempty(unvisited_nodes) && (numel(current_path) == n)
The first half of the if condition, isempty(unvisited_nodes) is standard. If you only use this part of the condition you'll get all paths from the start node to a leaf, regardless of path length. (Hence the cell array output.) The second half, (numel(current_path) == n) enforces the length of the path.
I took a shortcut here because n is the number of nodes in the adjacency matrix, which in the sample case is 8 rather than 7, the number of characters in your alphabet. But there are no edges into or out of node 1 because I was apparently planning on using a trick that I never got around to telling you about. Rather than run DFS starting from each of the nodes to get all of the paths, you can make a dummy node (in this case node 1) and create an edge from it to all of the other real nodes. Then you just call DFS once on node 1 and you get all the paths. Here's the updated adjacency matrix:
A =
0 1 1 1 1 1 1 1
0 0 1 0 0 1 0 0
0 1 0 0 0 0 0 1
0 0 0 0 0 1 0 1
0 1 0 0 0 0 1 0
0 1 0 0 0 0 1 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
If you don't want to use this trick, you can change the condition to n-1, or change the adjacency matrix not to include node 1. Note that if you do leave node 1 in, you need to remove it from the resulting paths.
Here's the output of the function using the updated matrix:
>> dft_all(A, 1)
ans =
{
[1,1] =
1 2 3 8 5 7 4 6
[1,2] =
1 3 2 6 7 4 8 5
[1,3] =
1 3 8 5 2 6 7 4
[1,4] =
1 3 8 5 7 4 6 2
[1,5] =
1 4 6 2 3 8 5 7
[1,6] =
1 5 7 4 6 2 3 8
[1,7] =
1 6 2 3 8 5 7 4
[1,8] =
1 6 7 4 8 5 2 3
[1,9] =
1 7 4 6 2 3 8 5
[1,10] =
1 8 5 7 4 6 2 3
}