In Tensorflow 1, if you run this code:
tfs = tf.InteractiveSession()
a = tf.Variable(5)
b = tf.constant(4)
tfs.run(tf.global_variables_initializer())
a.assign(b)
print(tfs.run(a))
It prints:
5
But it should print "4" instead of "5".
I cannot see why it's not working since the assign function is supposed to update the value of a.
Thank you
Related
I had the task to code the following:
Take a list of integers and returns the value of these numbers added up, but only if they are odd.
Example input: [1,5,3,2]
Output: 9
I did the code below and it worked perfectly.
numbers = [1,5,3,2]
print(numbers)
add_up_the_odds = []
for number in numbers:
if number % 2 == 1:
add_up_the_odds.append(number)
print(add_up_the_odds)
print(sum(add_up_the_odds))
Then I tried to re-code it using function definition / return:
def add_up_the_odds(numbers):
odds = []
for number in range(1,len(numbers)):
if number % 2 == 1:
odds.append(number)
return odds
numbers = [1,5,3,2]
print (sum(odds))
But I couldn’t make it working, anybody can help with that?
Note: I'm going to assume Python 3.x
It looks like you're defining your function, but never calling it.
When the interpreter finishes going through your function definition, the function is now there for you to use - but it never actually executes until you tell it to.
Between the last two lines in your code, you need to call add_up_the_odds() on your numbers array, and assign the result to the odds variable.
i.e. odds = add_up_the_odds(numbers)
I am drawing a plot of voltage per time. For the voltage values, I want the values to be evaluated by a 'scaling' function which converts the values from volts to kilovolts if the biggest element is higher than 1000 volts (11000 volts to 11 KILOvolts).
This function is supposed to return 2 separate outputs; one for (new) values of voltage and one for the unit. The values are fed into the y axis values of the plot and the unit is given to the labeling line of that axis. For example:
import numpy as np
time = np.array([0, 1, 2, 3])
system_voltage1 = np.array([110, 120, 130, 150])
system_voltage2 = np.array([11000, 12000, 13000, 15000])
scaling_function(input)
if np.amax(input) < 1000:
output = input/1
Voltage_label = 'Voltage in Volts'
if np.amax(input) > 1000:
output = input/1000
Voltage_label = 'Voltage in KILOVolts'
return(output, Voltage_label)
fig14 = plt.figure(figsize=(16,9))
ax1 = fig14.add_subplot(111)
l1, = ax1.plot(time, scaling_function(system_voltage), color='r')
ax1.set_xlabel("time in second", color='k')
ax1.set_ylabel(Voltage_label, color='k')
Now, I am having trouble, calling this function properly. I need the function to only receive the output for scaling_function(system_voltage), and receive Voltage_label in ax1.set_ylabel(Voltage_label, color='k'). Now:
A) My problem: I don't know how to write the code so only the first output is received and used for scaling_function(system_voltage) , and the second element for the labeling line.
B) Something I tried but didn't work:Voltage_label does not recognize the value of voltage_label from scaling_function, as it is located in an outer loop than the function. I mean, I cannot access voltage_label as its value is not globally assigned.
Can anyone help me with this?
y,l = scaling_function(system_voltage)
l1, = ax1.plot(time, y, color='r')
ax1.set_xlabel("time in second", color='k')
ax1.set_ylabel(l, color='k')
Until now, I change req manually. The code works, including saving the result into a file.
But now I want to run the code for all possible values of req.
Without saving it into a file, the code works but obviously it overwrite the result.
That is why I put that line of code that saving the result by giving it a different name depending of the values of req. But this gives me error.
error:
error: sprintf: wrong type argument 'cell'
error: called from
testforloop at line 26 column 1
my code:
clear all;
clc;
for req = {"del_1", "del_2", "del_3"}
request = req;
if (strcmp(request, "del_1"))
tarr = 11;
# and a bunch of other variables
elseif (strcmp(request, "del_2"))
tarr = 22;
# and a bunch of other variables
elseif (strcmp(request, "del_3"))
tarr = 33;
# and a bunch of other variables
else
# do nothing
endif
#long calculation producing many variable including aa, bb, cc.
aa = 2 * tarr;
bb = 3 * tarr;
cc = 4 * tarr;
#collecting variables of interest: aa, bb, cc and save it to a file.
result_matrix = [aa bb cc];
dlmwrite (sprintf('file_result_%s.csv', request), result_matrix);
endfor
if I use ["del_1" "del_2" "del_3"], the error is
error: 'tarr' undefined near line 20 column 10
error: called from
testforloop at line 20 column 4
Inside the loop
for req = {"del_1", "del_2", "del_3"}
req gets as value each of the cells of the cell array, not the contents of the cells (weird design decision, IMO, but this is the way it works). Thus, req={"del_1"} in the first iteration. The string itself can then be obtained with req{1}. So all you need to change is:
request = req{1};
However, I would implement this differently, as so:
function myfunction(request, tarr)
% long calculation producing many variable including aa, bb, cc.
aa = 2 * tarr;
bb = 3 * tarr;
cc = 4 * tarr;
% collecting variables of interest: aa, bb, cc and save it to a file.
result_matrix = [aa bb cc];
dlmwrite (sprintf('file_result_%s.csv', request), result_matrix);
end
myfunction("del_1", 11)
myfunction("del_2", 22)
myfunction("del_3", 33)
I think this obtains a clearer view of what you're actually doing, the code is less complicated.
Note that in Octave, ["del_1" "del_2" "del_3"] evaluates to "del_1del_2del_3". That is, you concatenate the strings. In MATLAB this is not the case, but Octave doesn't know the string type, and uses " in the same way as ' to create char arrays.
I am new to tensorflow framework. I am using
tf.reduce_logsumexp in my code. But inspecting the output I see that some of the values are negative. How is that possible? I suspected that it might be due to some nans or inf values so I put in a check to remove those values in my input like this(X is my input):
res = tf.where(tf.is_inf(X), tf.zeros_like(X), X)
res = tf.where(tf.is_nan(res), tf.zeros_like(res), res)
output = tf.reduce_logsumexp(res, axis=0)
But even this does not help and I still get some values as negative. Any help appreciated! Thanks
Note that the logarithm is negative, if its argument is smaller than 1. Therefore, you will always get a negative logsumexp output if the sum of the exponentials is smaller than 1. This occurs for example if all of the exponents are much smaller than zero, i.e. you have
res = [-2.5, -1.4, -3.3, -1.65, -2.15], then the corresponding exponentials are
exp(res) = [0.082, 0.247, 0.037, 0.192, 0.116], and their sum
sum(exp(res)) = 0.674 smaller than 1.
If you then take the logarithm, you get
log(sum(exp(res))) = log(0.674) = -0.394.
It simply emerges from the definition of the function that it does not necessarily have a positive output. You can test it with the following program:
import numpy as np
def logsumexp(arr):
summ = 0.0
for i in range(arr.shape[0]):
print(np.exp(arr[i]))
summ += np.exp(arr[i])
print('Sum: {}'.format(summ))
return np.log(np.sum(np.exp(arr)))
arr = np.asarray([-2.5, -1.4, -3.3, -1.65, -2.15])
print('LogSumExp: {}'.format(logsumexp(arr)))
As a homework assignment, I'm writing a code that uses the bisection method to calculate the root of a function with one variable within a range. I created a user function that does the calculations, but one of the inputs of the function is supposed to be "fun" which is supposed to be set equal to the function.
Here is my code, before I go on:
function [ Ts ] = BisectionRoot( fun,a,b,TolMax )
%This function finds the value of Ts by finding the root of a given function within a given range to a given
%tolerance, using the Bisection Method.
Fa = fun(a);
Fb = fun(b);
if Fa * Fb > 0
disp('Error: The function has no roots in between the given bounds')
else
xNS = (a + b)/2;
toli = abs((b-a)/2);
FxNS = fun(xns);
if FxNS == 0
Ts = xNS;
break
end
if toli , TolMax
Ts = xNS;
break
end
if fun(a) * FxNS < 0
b = xNS;
else
a = xNS;
end
end
Ts
end
The input arguments are defined by our teacher, so I can't mess with them. We're supposed to set those variables in the command window before running the function. That way, we can use the program later on for other things. (Even though I think fzero() can be used to do this)
My problem is that I'm not sure how to set fun to something, and then use that in a way that I can do fun(a) or fun(b). In our book they do something they call defining f(x) as an anonymous function. They do this for an example problem:
F = # (x) 8-4.5*(x-sin(x))
But when I try doing that, I get the error, Error: Unexpected MATLAB operator.
If you guys want to try running the program to test your solutions before posting (hopefully my program works!) you can use these variables from an example in the book:
fun = 8 - 4.5*(x - sin(x))
a = 2
b = 3
TolMax = .001
The answer the get in the book for using those is 2.430664.
I'm sure the answer to this is incredibly easy and straightforward, but for some reason, I can't find a way to do it! Thank you for your help.
To get you going, it looks like your example is missing some syntax. Instead of either of these (from your question):
fun = 8 - 4.5*(x - sin(x)) % Missing function handle declaration symbol "#"
F = # (x) 8-4.5*(x-sin9(x)) %Unless you have defined it, there is no function "sin9"
Use
fun = #(x) 8 - 4.5*(x - sin(x))
Then you would call your function like this:
fun = #(x) 8 - 4.5*(x - sin(x));
a = 2;
b = 3;
TolMax = .001;
root = BisectionRoot( fun,a,b,TolMax );
To debug (which you will need to do), use the debugger.
The command dbstop if error stops execution and opens the file at the point of the problem, letting you examine the variable values and function stack.
Clicking on the "-" marks in the editor creates a break point, forcing the function to pause execution at that point, again so that you can examine the contents. Note that you can step through the code line by line using the debug buttons at the top of the editor.
dbquit quits debug mode
dbclear all clears all break points