I have a rather specific error in netlogo that I've been staring at for a while now. Hope you guys have some insight.
The error is in a code which looks back in a list called 'strategy'. If the list is longer than investment-time variables 'REfocus' and 'PRICE' are set to a certain value. If the list is not longer than investment-time, the variables are not set (and thus remain 0).
The code consists out of a function strategy_actions and a reporter investment_time. Investment-time is approximately 3 years, but as ticks are in months, investment-time is rescaled to months. In strategy_actions, investment-time is scaled back to years, as each entry in the strategy list is also annual. (The scaling and rescaling seems arbitrary, but as investment-time is used a lot by other parts of the code, it made more sense to do it like this). The goal is to take the strategy from x time back (equal to investment-time).
The code (error follows underneath):
to strategy_actions
set_ROI
start_supply?
if current_strategy != 0
[
let it (investment_time / 12)
ifelse it >= length strategy
[
set REfocus 0
]
[
if item (it - 1) strategy = 1
[
if supply? = true [set_PRICE (set_discrete_distribution 0.29 0.19 0.29 0.15 0.07 0 0) (set_discrete_distribution 0.14 0.12 0.25 0.25 0.25 0 0)]
ifelse any? ids [set REfocus mean [mot_RE] of ids][set REfocus set_discrete_distribution 0.07 0.03 0.07 0.17 0.66 0 0]
]
if item (it - 1) strategy = 2
[
if supply? = true [set_PRICE (set_discrete_distribution 0.27 0.21 0.32 0.11 0.09 0 0) (set_discrete_distribution 0.15 0.11 0.22 0.30 0.23 0 0)]
ifelse any? prods [set REfocus mean [mot_RE] of prods][set REfocus set_discrete_distribution 0.12 0.03 0.10 0.18 0.57 0 0]
]
if item (it - 1) strategy = 3
[
if supply? = true [set_PRICE (set_discrete_distribution 0.26 0.22 0.26 0.18 0.09 0 0) (set_discrete_distribution 0.07 0.08 0.19 0.30 0.35 0 0)]
ifelse any? cons[set REfocus mean [mot_RE] of cons][set REfocus set_discrete_distribution 0.08 0.06 0.15 0.27 0.45 0 0]
]
]
set RE_history fput REfocus RE_history
]
end
to-report investment_time
report ((random-normal 3 1) * 12) ;approximately 3 years investment time
end
somehow, i sometimes get this runtime error during my behaviorspace experiment:
-1 isn't greater than or equal to zero.
error while observer running ITEM
called by procedure STRATEGY_ACTIONS
called by procedure SET_MEETING_ACTIONS
called by procedure GO
Does anyone know what causes this error?
You would help me out a lot!
Cheers,
Maria
It appears that investment_time is occasionally coming in as zero, so you are asking for item (0 - 1) of the strategy list. I did a bit of playing around with item and learned (to my surprise) that item (0.0001 - 1) strategy works just fine, yielding the 0th item in the list in spite of the argument being negative. But item (0 - 1) strategy does give the error you cite. Apparently an item number greater than -1 is interpreted as zero. Indeed item seems to truncate any fractional argument rather than rounding it. E.g., item 0.9 is interpreted as item 0, as is item -0.9
That might be worth putting in the documentation.
HTH,
Charles
Related
I have a list of observations with a few variables. I need to put them in a bin (below) and only keep one observation in each bin which is closest to the bin's number:
Bins
0.94
0.96
0.98
1.00
1.02
1.04
1.06
Data
Variable Price Value_to_bin Closest bin
a 0.630527682 0.935 0.94
b 0.441296291 0.979 0.98
c 0.350173415 0.969
d 0.920932417 0.993
e 0.361863025 0.959 0.96
f 0.027205755 1.003 1
g 0.878286791 1.045
h 0.206434946 0.971
i 0.259272294 1.021 1.02
j 0.081774863 0.982
k 0.01146324 0.992
l 0.283027273 1.037 1.04
m 0.188747537 0.993
n 0.554786 1.064 1.06
o 0.784774 1.065
And then just keep the ones that are closest to the bin value (i.e. delete the ones that have blanks in the 'closest_bin' variable.
I tried to use proc rank but I can't get rid of the rest or match with the bin (something like 'closest' doesn't exist as far as I know).
SAS SQL with automatic remerging can perform the query quite succinctly. The consistent binning to a 0.02 level allows the ROUND function to compute the bin values to the nearest 0.02 unit.
proc sql;
create table want as
select
var,
price,
value,
round(value,0.02) as valbin_02
from have
group by valbin_02
having abs(valbin_02-value) = min(abs(valbin_02-value))
;
I have a result from the classification_report from sklearn.metrics and then print the report it would be like:
precision recall f1-score support
1 1.00 0.84 0.91 43
2 0.12 1.00 0.22 1
avg / total 0.98 0.84 0.90 44
Now, the question is how can I show the result in a Jupyter widget (in the above format) and update its value?
Currently, I am using html widgets to show the result:
#pass test and result vectors
report = classification_report(pred_test , self.y_test_data)
predict_table = widgets.HTML(value = "")
predict_table.value = report
but it likes the following:
precision recall f1-score support 1 1.00 0.81 0.90 43 2 0.00 0.00 0.00 0 avg / total 1.00 0.81 0.90 43
I found a simple solution using html techniques! As we are using html widget in Jupyter, the problem can be solved by using pre tag in html:
predict_table.value = "<pre>" + report + "</pre>
I have a large pandas dataframe with 2000 rows (one date per row) and 2000 columns (1 second intervals). Each cell represents a temperature reading.
Starting with the 5th row, I need to go back 5 rows and find all the observations where the the 1st column in the row is higher than the 2nd column in the row.
For the 5th row I may find 2 such observations. I then want to do summary stats on the observations and append those summary stats to a list.
Then I go to the 6st row and go back 5 rows and find all the obvs where the 1th column is higher than the 2nd column. I get all obvs, do summary stats on the obvs and append the results to the new dataframe.
So for each row in the dataframe, I want to go back 5 days, get the obvs, get the stats, and append the stats to a dataframe.
The problem is that if I perform this operation on rows 5 -2000, then I will have a for-loop that is 1995 cycles long, and this takes a while.
What is the better or best way to do this?
Here is the code:
print huge_dataframe
sec_1 sec_2 sec_3 sec_4 sec_5
2013_12_27 0.05 0.12 0.06 0.15 0.14
2013_12_28 0.06 0.32 0.56 0.14 0.17
2013_12_29 0.07 0.52 0.36 0.13 0.13
2013_12_30 0.02 0.12 0.16 0.55 0.12
2013_12_31 0.06 0.30 0.06 0.14 0.01
2014_01_01 0.05 0.12 0.06 0.15 0.14
2014_01_02 0.06 0.32 0.56 0.14 0.17
2014_01_03 0.07 0.52 0.36 0.13 0.13
2014_01_04 0.02 0.12 0.16 0.55 0.12
2014_01_05 0.06 0.30 0.06 0.14 0.01
for each row in huge_dataframe.ix[5:]:
move = row[sec_1] - row[sec_2]
if move < 0: move = 'DOWN'
elif move > 0: move = 'UP'
relevant_dataframe = huge_dataframe.ix[only the 5 rows preceding the current row]
if move == 'UP':
mask = relevant_dataframe[sec_1 < sec_2] # creates a boolean dataframe
observations_df = relevant_dataframe[mask]
elif move == 'DOWN':
mask = relevant_dataframe[sec_1 > sec_2] # creates a boolean dataframe
observations_df = relevant_dataframe[mask]
# At this point I have observations_df which is only filled
# with rows where sec_1 < sec_2 or the opposite, depending on which
# row I am in.
summary_stats = str(observations_df.describe())
summary_list.append(summary_stats) # This is the goal
# I want to ultimatly
# turn the list into a
# dataframe
Since there is no code to create the data, I will just sketch the code that I would try to make work. Generally, try to prevent from row-wise operations whenever you can. I first had no clue either, but then I got interested and some research yielded TimeGrouper:
df = big_dataframe
df['move'] = df['sec_1'] > df['sec2']
def foobarRules(group):
# keep in mind that in here, you refer not to "relevant_dataframe", but to "group"
if (group.tail(1).move == True):
# some logic
else:
# some other logic
return str(group.describe())
grouper = TimeGrouper('5D')
allMyStatistics = df.groupby(grouper).apply(foobarRules)
I have honestly no clue how the return works if you return a multi-dimensional dataframe. I know it works well if you return either a row or a column, but if you return a dataframe that contains both rows and columns for every group - I guess pandas is smart enough to compute a panel of all these. Well, you will find out.
My original observations look like that:
name Analyte
spring 0.1
winter 0.4
To calculate p-value I did bootstrapping simulation:
name Analyte
spring 0.001
winter 0
spring 0
winter 0.2
spring 0.03
winter 0
spring 0.01
winter 0.02
spring 0.1
winter 0.5
spring 0
winter 0.04
spring 0.2
winter 0
spring 0
winter 0.06
spring 0
winter 0
.....
Now I want to calculate empirical p-value: In original data winter Analyte = 0.4 - if in bootstrapped data winter analyte was >=0.4 (for example 1 time) and bootstrapping was done (for example 100 times), then empirical p-value for winter analyte is calculated:
1/100 = 0.01
(How many times data was the same or higher than in original data
divided by total number of observations)
For spring analyte p-value is:
2/100 = 0.02
I want to calculate those p-values with awk.
My solution for spring is:
awk -v VAR="spring" '($1==VAR && $2>=0.1) {n++} END {print VAR,"p-value=",n/100}'
spring p-value= 0.02
The help I need is to pass original file (with names spring and winter and their analytes, observations and number of observations) into awk and assign those.
Explanation and script content:
Run it like: awk -f script.awk original bootstrap
# Slurp the original file in an array a
# Ignore the header
NR==FNR && NR>1 {
# Index of this array will be type
# Value of that type will be original value
a[$1]=$2
next
}
# If in the bootstrap file value
# of second column is greater than original value
FNR>1 && $2>a[$1] {
# Increment an array indexed at first column
# which is nothing but type
b[$1]++
}
# Increment another array regardless to identify
# the number of times bootstrapping was done
{
c[$1]++
}
# for each type in array a
END {
for (type in a) {
# print the type and calculate empirical p-value
# which is done by dividing the number of times higher value
# of a type was seen and total number of times
# bootstrapping was done.
print type, b[type]/c[type]
}
}
Test:
$ cat original
name Analyte
spring 0.1
winter 0.4
$ cat bootstrap
name Analyte
spring 0.001
winter 0
spring 0
winter 0.2
spring 0.03
winter 0
spring 0.01
winter 0.02
spring 0.1
winter 0.5
spring 0
winter 0.04
spring 0.2
winter 0
spring 0
winter 0.06
spring 0
winter 0
$ awk -f s.awk original bootstrap
spring 0.111111
winter 0.111111
Analysis:
Spring Original Value is 0.1
Winter Original Value is 0.4
Bootstrapping done is 9 times for this sample file
Count of values higher than Spring original value = 1
Count of values higher than Winter's original value = 1
So, 1/9 = 0.111111
this works for me, (GNU awk 3.1.6):
FNR == NR {
a[$1] = $2
next
}
$2 > a[$1] {
b[$1]++
}
{
c[$1]++
}
END {
for (i in a) print i, "p-value=",b[i]/c[i]
}
..output is:
winter p-value= 0.111111
spring p-value= 0.111111
I'm working this programming task for college where we have to write a c++ program that calculates the magnetic field vector for certain coils in 3D space.
I've managed to write this program and I think I've got it working pretty well.
I want to add in a special thinh though (it's my exam paper, so it has to be extra good!): I wan't to plot the vectors out.
I'm used to calling gnuplot from c++ (via piping) and this is what I usually do:
create an output stream that writes the data to a .dat file
open a gnuplot pipe
make gnuplot plot all the contents of the .dat
Since my data has always been 2D, xand y plots, I'm quite lost here. My question is:
How to format the .dat file (e.g. do I use braces to group vector components?)
what is the actual gnuplot command to plot a 3D vector field?
It'd be easy if I could format the .dat file like this:
# Px Py Pz Bx By Bz
1 0 2 0.7 0.5 0.25 #<= example data line
... more data ...
when the magnetic field vector in the point P=(1,0,2)equals a vector B=(0.7,0.5,0.25). This would be easy to program, the real question is: will this do ? and how to I plot it in gnuplot. (wow, I've asked the same question 3 times I guess).
Piping to gnuplot
Ok, since someone asked me to describe how I pipe (don't know if it's the right term thought) stuff to gnuplot. Here it is:
First open up a pipe and call it pipe:
FILE *pipe = popen("gnuplot -persist 2>/dev/null", "w");
Tell gnuplot what to do through the pipe:
fprintf(pipe, "set term x11 enhanced \n");
fprintf(pipe, "plot x^2 ti 'x^2' with lines\n");
notice the \nwhich is absolutely necessary. It is what executes the command.
close the pipe:
pclose(pipe);
The necessary library is called <fstream> I believe.
I made this simple example to show you how to draw a vector field. The output would be something like this pic:
The data example I used to plot this was:
# Px Py Pz Bx By Bz
0 0 0 0.8 0.8 0.45
0 0 1 0.5 0.7 0.35
0 0 2 0.7 0.5 0.25
0 1 0 0.65 0.65 0.50
0 1 1 0.6 0.6 0.3
0 1 2 0.45 0.45 0.20
1 0 0 0.5 0.7 0.35
1 0 1 0.75 0.75 0.4
1 0 2 0.85 0.85 0.25
1 1 0 0.90 0.85 0.23
1 1 1 0.95 0.86 0.20
1 1 2 0.98 0.88 0.13
2 0 0 0.73 0.83 0.43
2 0 1 0.53 0.73 0.33
2 0 2 0.73 0.53 0.23
2 1 0 0.68 0.68 0.52
2 1 1 0.63 0.57 0.23
2 1 2 0.48 0.42 0.22
The command to plot it is:
gnuplot> splot "./data3d.dat" with vectors
Now you should read the section 44, page 53 of the official manual (and here the pdf). You may find this site also very useful.
Edited:
This command doesn't fit into your description: mapping from (x,y,z) to (t,u,v). It actually does this mapping: from (X,Y,Z) to (X+dX,Y+dY,Z+dZ).
Cheers,
Beco