Consider the following input in a IPython notebook:
mu = 39.95
Z=1
Latex(r"Atomic Mass: $\quad \mu$= %0.5e Hz" %(mu))
Latex(r"Charge state: $\quad z$= %0.5e" %(Z))
My question has two parts.
(A) Missing output
From above source, Surprisingly I got the following result:
Charge state: z = 1.00000e+00
What happened to the first line (i.e the value of $mu$ was not printed)?
(B) Missing newline
I was however able to get the result using the following:
Latex(r"Atomic Mass: $\quad \mu$= %0.5e Hz " %(mu) + r"Charge state: $\quad z$= %0.5e" %(Z))
But now, I need a newline in the above. How does one do it?
By default the notebook only renders the last value obtained in a cell. Use IPython.display.display as an augmented print function to display several things:
from IPython.display import Latex, display
mu = 39.95
Z=1
display(Latex(r"Atomic Mass: $\quad \mu$= %0.5e Hz" % mu))
display(Latex(r"Charge state: $\quad z$= %0.5e" % Z))
Related
I'm very new to caret package and nnet in R. I've done some projects related to ANN with Matlab before, but now I need to work with R and I need some basic help.
My input dataset has 1000 observations (in rows) and 23 variables (in columns). My output has 1000 observations and 12 variables.
Here are some sample data that represent my dataset and might help to understand my problem better:
input = as.data.frame(matrix(sample(1 : 20, 100, replace = TRUE), ncol = 10))
colnames(input) = paste ( "X" , 1:10, sep = "") #10 observations and 10 variables
output = as.data.frame(matrix(sample(1 : 20, 70, replace = TRUE), ncol = 7))
colnames(output) = paste ( "Y" , 1:7, sep = "") #10 observations and 7 variables
#nnet with caret:
net1 = train(output ~., data = input, method= "nnet", maxit = 1000)
When I run the code, I get this error:
error: invalid type (list) for variable 'output'.
I think I have to add all output variables separately (which is very annoying, especially with a lot of variables), like this:
train(output$Y1 + output$Y2 + output$Y3 + output$Y4 + output$Y5 +
output$Y6 + output$Y7 ~., data = input, method= "nnet", maxit = 1000)
This time it runs but I get this error:
Error in [.data.frame(data, , all.vars(Terms), drop = FALSE) :
undefined columns selected
I try to use neuralnet package, with the code below it works perfectly but I still have to add output variables separately :(
net1 = neuralnet(output$Y1 + output$Y2 + output$Y3 + output$Y4 +
output$Y5 + output$Y6 + output$Y7 ~., data = input, hidden=c(2,10))
p.s. since these sample data are created randomly, the neuralnet cannot converge, but in my real data it works well (in comparison to Matlab ANN)
Now, if you could help me with a way to put output variables automatically (not manually), it solves my problem (although with neuralnet not caret).
use the str() function and ascertain that its a data frame looks like you are inputting a list to the train function. This may be because of a transformation you are doing before to output.
str(output)
Without a full script of earlier steps its difficult to understand what is going on.
After trying different things and searches, I finally found a solution:
First, we must use as.formula to show the relation between our input and output. With the code below we don't need to add all the variables separately:
names1 <- colnames(output) #the name of our variables in the output
names2 = colnames(input) #the name of our variables in the input
a <- as.formula(paste(paste(names1,collapse='+', sep = ""),' ~ '
,paste(names2,collapse='+', sep = "")))
then we have to combine our input and output in a single data frame:
all_data = cbind(output, input)
then, use neuralnet like this:
net1 = neuralnet(formula = a, data = all_data, hidden=c(2,10))
plot(net1)
This is also work with the caret package:
net1 = train(a, data = all_data, method= "nnet", maxit = 1000)
but it seems neuralnet works faster (at least in my case).
I hope this helps someone else.
Take this simple networkx script.
import networkx as nx
g = nx.DiGraph()
g.add_edge("a","b")
g.add_edge("a","c")
g.add_node("D")
dotter = nx.nx_agraph.to_agraph(g)
dotter.layout("dot")
dot = dotter.to_string()
with open("test_w_pos.dot","w") as fo:
fo.write(dot)
The output is:
strict digraph "" {
graph [bb="0,0,162,108"];
node [label="\N"];
a [height=0.5,
pos="63,90",
width=0.75];
b [height=0.5,
pos="27,18",
width=0.75];
a -> b [pos="e,35.634,35.269 54.285,72.571 50.04,64.081 44.846,53.693 40.134,44.267"];
c [height=0.5,
pos="99,18",
width=0.75];
a -> c [pos="e,90.366,35.269 71.715,72.571 75.96,64.081 81.154,53.693 85.866,44.267"];
D [height=0.5,
pos="135,90",
width=0.75];
}
I'd like to not have any of the pos, height and width attributes as they vary from run to run and make it hard to compare to previous runs on complex graphs.
desired output
strict digraph "" {
node [label="\N"];
a;
b;
a -> b;
c;
a -> c;
D;
}
Both dot files give substantially the same output in svg (see below), so I find the positioning unnecessary and in fact I've never seen this information put into dot files previously.
I could not find anything in to_agraph or dotter.to_string, but I did not notice extra arguments are possible in dotter.layout(dot):
agraph.py
def layout(self, prog="neato", 👉args=""👈):
"""Assign positions to nodes in graph.
Yes, with this very simplistic file, I could drop the positioning data afterwards. But in a real graph, there will be a lot of other color and label attributes mixed in with the positions.
image with position attributes:
image without position attributes:
If you are only interested in the dot file you can use the pydot interface of networkx and call write_dot, e.g.
drawing.nx_pydot.write_dot(g, "test.dot")
Try dotter.layout("canon") (https://graphviz.org/docs/outputs/canon/). This will provide a (pretty printed) version of your graph without any positioning.
I'd like to write several messages and tables on the same .txt file.
For example:
x=[23.9,10.9,8.9,14.2]
y=[9.83,8.04,7.47,8.32]
file=fopen('Results.txt','wt');
fprintf(file,'Results1\n');
fprintf(file,'%.2f %.2f\r\n',x,y);
fprintf(file,'Results2\n');
fclose(file);
I get this result as .txt:
Results1
23.90 10.90
8.90 14.20
9.83 8.04
7.47 8.32
Results2
But I should get this one:
Results1
23.90 9.83
10.90 8.04
8.90 7.47
14.20 8.32
Results2
Instead of fprintf(file,'%.2f %.2f\r\n',x,y);), I was trying to use:
ResultsTable2 = table(x,y);
writetable(file,ResultsTable2);
but didn't succeed. How to write the required .txt file?
Careful examination of your output shows that all the elements of x were printed before all the elements of y.
The documentation confirms that this is the expected behavior. Check out this example
A1 = [9.9, 9900];
A2 = [8.8, 7.7 ; ...
8800, 7700];
formatSpec = 'X is %4.2f meters or %8.3f mm\n';
fprintf(formatSpec,A1,A2)
X is 9.90 meters or 9900.000 mm
X is 8.80 meters or 8800.000 mm
X is 7.70 meters or 7700.000 mm
Even though the arguments to fprintf are in the order A1, A2. It first prints all the values from A1, and then it prints all the values from A2 going in single index order.
Therefore, if you want to alternate values from x and y during printing, you need to interleave them in a new variable. There are several possible ways to do so.
One example,
XY = reshape([x;y], 1, []);
Then everything should print as expected
fprintf(file, '%.2f %.2f\r\n', XY);
% or if you want to print to command window
% fprintf('%.2f %.2f\r\n', XY);
23.90 9.83
10.90 8.04
8.90 7.47
14.20 8.32
The correct answer for how to output data with fprintf is given by Cecilia: each argument will be iterated completely through in the order it appears in the argument list, so you have to combine the data into one matrix argument that will be iterated through column-wise to generate the desired output.
You also mentioned trying to use a table and the writetable function, so I though I'd add the correct way to do that in case you were curious:
ResultsTable2 = table(x(:), y(:)); % Pass data as column vectors
writetable(ResultsTable2, 'Results.txt', 'WriteVariableNames', false);
I'm new to the SPSS macro syntax and had a hard time trying to label variables based on a simple loop counter. Here's what I tried to do:
define !make_indicatorvars()
!do !i = 1 !to 10.
!let !indicvar = !concat('indexvar_value_', !i, '_ind')
compute !indicvar = 0.
if(indexvar = !i) !indicvar = 1.
variable labels !indicvar 'Indexvar has value ' + !quote(!i).
value labels !indicvar 0 "No" 1 "Yes".
!doend
!enddefine.
However, when I run this, I get the following warnings:
Warning # 207 on line ... in column ... Text: ...
A '+' was found following a text string, indicating continuation, but the next non-blank character was not a quotation mark or an apostrophe.
Warning # 4465 in column ... Text: ...
An invalid symbol appears on the VAR LABELS command where a slash was
expected. All text through the next slash will be be ignored.
Indeed the label is then only 'Indexvar has value '.
Upon using "set mprint on printback on", the following code was printed:
variable labels indexvar_value_1_ind 'Indexvar has value ' '1'
So it appears that SPSS seems to somehow remove the "+" which is supposed to concatenate the two strings, but why?
The rest of the macro worked fine, it's only the variable labels command that's causing problems.
Try:
variable labels !indicvar !quote(!concat('Indexvar has value ',!i)).
Also note:
compute !indicvar = 0.
if(indexvar = !i) !indicvar = 1.
Can be simplified as:
compute !indicvar = (indexvar = !i).
Where the right hand side of the COMPUTE equation evaluates to equal TRUE a 1 (one) is assigned else if FALSE a 0 (zero) is assigned. Using just a single compute in this way not only reduce the lines of code, it will also make the transformations more efficient/quicker to run.
You might consider the SPSSINC CREATE DUMMIES extension command. It will automatically construct a set of dummies for a variable and label them with the values or value labels. It also creates a macro that lists all the variables. There is no need to enumerate the values. It creates dummies for all the values in the data.
It appears on the Transform menu as long as the Python Essentials are installed. Here is an example using the employee data.sav file shipped with Statistics.
SPSSINC CREATE DUMMIES VARIABLE=jobcat
ROOTNAME1=job
/OPTIONS ORDER=A USEVALUELABELS=YES USEML=YES OMITFIRST=NO
MACRONAME1="!jobcat".
It looks like this has been asked many times, but none of the past posts seem to solve my question. All those had to do with matrix/vector while my code does not have any of these, just simple variables. It takes three variables as arguments. It works perfectly fine within the Matlab environment. I only got the error when I compiled it with mcc -m Normal.m and tried to run with the executable like this "./Normal 1 5 0.5". The complete error message is:
Error using /
Matrix dimensions must agree.
Error in Normal (line 4)
MATLAB:dimagree
It is complaining about line 4: N=2/dt, what is wrong with this?
Here is the code:
function val=Normal(l1,l2,dt)
const=(l2/l1-1);
N=2/dt;
S=1.0/sqrt(l2/l1);
Z(1)=S;
for i=2:N
t= -1+(i-1)*dt;
Z(i)=1.0/sqrt(const*t*t+1);
S=S+2*Z(i);
end
Z(21)=1.0/(l2/l1);
S=S+1.0/sqrt(l2/l1);
val=dt*S/2;
end
But dt is not a scalar when passed into the standalone through the command ./Normal 1 5 0.5. It is a character array with 3 elements ('0', '.','5')!
When passing numerical arguments to a standalone, they are passed as strings. Thus, inside the function, you need to convert '0.5' into a double, and similarly for l1 and l2:
dt = str2num(dt);
l1 = str2num(l1);
l2 = str2num(l2);
Note that you can use isdeployed to determine at runtime if the function is a standalone:
if isdeployed, dt = str2num(dt); end
And you might need to display the result:
if isdeployed, disp(val); end
Result:
>> system('Normal 1 5 0.5');
1.4307
>> Normal(1,5,0.5) % .m function for comparison
ans =
1.4307