I am working with a educational dataset called IPEDS from the National Center for Educational Statistics. They track students in college based upon major, degree completion, etc. The problem in Stata is that I am trying to determine the total count for degrees obtained by a specific major.
They have a variable cipcode which contains values that serve as "majors". cipcode might be 14.2501 "petroleum engineering, 16.0102 "Linguistics" and so forth.
When I write a particular code like
tab cipcode if cipcode==14.2501
it reports no observations. What code will give me the totals?
/*Convert Float Variable to String Variable and use Force Replace*/
tostring cipcode, gen(cipcode_str) format(%6.4f) force
replace cipcode_str = reverse(substr(reverse(cipcode_str), indexnot(reverse(cipcode_str), "0"), .))
replace cipcode_str = reverse(substr(reverse(cipcode_str), indexnot(reverse(cipcode_str), "."), .))
/* Created a total variable called total_t1 for total count of all stem majors listed in table 1*/
gen total_t1 = cipcode_str== "14.2501" + "14.3901" + "15.0999" + "40.0601"
This minimal example confirms your problem. (See, by the way, https://stackoverflow.com/help/mcve for advice on good examples.)
* code
clear
input code
14.2501
14.2501
14.2501
end
tab code if code == 14.2501
tab code if code == float(14.2501)
* results
. tab code if code == 14.2501
no observations
. tab code if code == float(14.2501)
code | Freq. Percent Cum.
------------+-----------------------------------
14.2501 | 3 100.00 100.00
------------+-----------------------------------
Total | 3 100.00
The keyword is one you use, precision. In Stata, search precision for resources, starting with blog posts by William Gould. A decimal like 14.2501 is hard (impossible) to hold exactly in binary and the details of holding a variable as type float can bite.
It's hard to see what you're doing with your last block of code, which you don't explain. The last statement looks puzzling, as you're adding strings. Consider what happens with
. gen whatever = "14.2501" + "14.3901" + "15.0999" + "40.0601"
. di whatever[1]
14.250114.390115.099940.0601
The result is a long string that cannot be a valid cipcode. I suspect that you are reaching towards
... if inlist(cipcode_str, "14.2501", "14.3901", "15.0999", "40.0601")
which is quite different.
But using float() is the minimal trick for this problem.
Related
I have confusion about the numbers at the end of the branches of a J48 tree. For example, using the weather.nominal data the tree looks the same, whether the Test options are set to Use training set or Cross-validation or Percentage split.
This is the output:
J48 pruned tree
------------------
outlook = sunny
| humidity = high: no (3.0)
| humidity = normal: yes (2.0)
outlook = overcast: yes (4.0)
outlook = rainy
| windy = TRUE: no (2.0)
| windy = FALSE: yes (3.0)
According to the textbook by the authors of this software, in an example using this exact data they say, "In the tree structure, a colon introduces the class label that has been assigned to a particular leaf, followed by the number of instances that reach that leaf, expressed as a decimal number because of the way the algorithm uses fractional instances to handle missing values. If there were incorrectly classified instances (there aren’t in this example) their number would appear, too: thus 2.0/1.0 means that two instances reached that leaf, of which one is classified incorrectly"
So this means that no instances were incorrectly classified in the above tree with the weather.nominal dataset.
On the other hand, when the test options are set to either 'Use training set' or 'Percentage split' (with the default random seed), there are incorrectly classified instances. For example, with a 60 percentage split, it shows the following
=== Evaluation on test split ===
=== Summary ===
Correctly Classified Instances 2 40 %
Incorrectly Classified Instances 3 60 %
There seems to be a contradiction here but I must be missing something. Is the tree shown initially not the tree that is built with the 60 percentage split?
That is not stated anywhere as far as I have seen but I can't think of any other explanation.
Just for completeness, the data is here:
outlook,temperature,humidity,windy,play
sunny,hot,high,FALSE,no
sunny,hot,high,TRUE,no
overcast,hot,high,FALSE,yes
rainy,mild,high,FALSE,yes
rainy,cool,normal,FALSE,yes
rainy,cool,normal,TRUE,no
overcast,cool,normal,TRUE,yes
sunny,mild,high,FALSE,no
sunny,cool,normal,FALSE,yes
rainy,mild,normal,FALSE,yes
sunny,mild,normal,TRUE,yes
overcast,mild,high,TRUE,yes
overcast,hot,normal,FALSE,yes
rainy,mild,high,TRUE,no
If you take a closer look at the output, you will see the following:
=== Classifier model (full training set) ===
The model that is being depicted there is the model that was trained on the full dataset, not your split.
The next section has the following heading:
=== Evaluation on test split ===
The statistics that you are referring to are based on a model trained and evaluated on your dataset split.
I am trying to run a multinomial logit with year fixed effects in mlogit in Stata (panel data: year-country), but I do not get standard errors for some of the models. When I run the same model using multinom in R I get both coefficients and standard errors.
I do not use Stata frequently, so I may be missing something or I may be running different models in Stata and R and should not be comparing them in the first place. What may be happening?
So a few details about the simple version of the model of interest:
I created a data example to show what the problem is
Dependent variable (will call it DV1) with 3 categories of -1, 0, 1 (unordered and 0 as reference)
Independent variables: 2 continuous variables, 3 binary variables, interaction of 2 of the 3 binary variables
Years: 1995-2003
Number of observations in the model: 900
In R I run the code and get coefficients and standard errors as below.
R version of code creating data and running the model:
## Fabricate example data
library(fabricatr)
data <- fabricate(
N = 900,
id = rep(1:900, 1),
IV1 = draw_binary(0.5, N = N),
IV2 = draw_binary(0.5, N = N),
IV3 = draw_binary(0.5, N = N),
IV4 = draw_normal_icc(mean = 3, N = N, clusters = id, ICC = 0.99),
IV5 = draw_normal_icc(mean = 6, N = N, clusters = id, ICC = 0.99))
library(AlgDesign)
DV = gen.factorial(c(3), 1, center=TRUE, varNames=c("DV"))
year = gen.factorial(c(9), 1, center=TRUE, varNames=c("year"))
DV = do.call("rbind", replicate(300, DV, simplify = FALSE))
year = do.call("rbind", replicate(100, year, simplify = FALSE))
year[year==-4]= 1995
year[year==-3]= 1996
year[year==-2]= 1997
year[year==-1]= 1998
year[year==0]= 1999
year[year==1]= 2000
year[year==2]= 2001
year[year==3]= 2002
year[year==4]= 2003
data1=cbind(data, DV, year)
data1$DV1 = relevel(factor(data1$DV), ref = "0")
## Save data as csv file (to use in Stata)
library(foreign)
write.csv(data1, "datafile.csv", row.names=FALSE)
## Run multinom
library(nnet)
model1 <- multinom(DV1 ~ IV1 + IV2 + IV3 + IV4 + IV5 + IV1*IV2 + as.factor(year), data = data1)
Results from R
When I run the model using mlogit (without fixed effects) in Stata I get both coefficients and standard errors.
So I tried including year fixed effects in the model using Stata three different ways and none worked:
femlogit
factor-variable and time-series operators not allowed
depvar and indepvars may not contain factor variables or time-series operators
mlogit
fe option: fe not allowed
used i.year: omits certain variables and/or does not give me standard errors and only shows coefficients (example in code below)
* Read file
import delimited using "datafile.csv", clear case(preserve)
* Run regression
mlogit DV1 IV1 IV2 IV3 IV4 IV5 IV1##IV2 i.year, base(0) iterate(1000)
Stata results
xtmlogit
error - does not run
error message: total number of permutations is 2,389,461,218; this many permutations require a considerable amount of memory and can result in long run times; use option force to proceed anyway, or consider using option rsample()
Fixed effects and non-linear models (such as logits) are an awkward combination. In a linear model you can simply add dummies/demean to get rid of a group-specific intercept, but in a non-linear model none of that works. I mean you could do it technically (which I think is what the R code is doing) but conceptually it is very unclear what that actually does.
Econometricians have spent a lot of time working on this, which has led to some work-arounds, usually referred to as conditional logit. IIRC this is what's implemented in femlogit. I think the mistake in your code is that you tried to include the fixed effects through a dummy specification (i.year). Instead, you should xtset your data and then run femlogit without the dummies.
xtset year
femlogit DV1 IV1 IV2 IV3 IV4 IV5 IV1##IV2
Note that these conditional logit models can be very slow. Personally, I'm more a fan of running two one-vs-all linear regressions (1=1 and 0/-1 set to zero, then -1=1 and 0/1 set to zero). However, opinions are divided (Wooldridge appears to be a fan too, many others very much not so).
Following my earlier question, I have tried to work on a code to return a string if a search term in a certain list is in a string to be returned as follows.
import re
from nltk import tokenize
from nltk.tokenize import sent_tokenize
def foo():
List1 = ['risk','cancer','ocp','hormone','OCP',]
txt = "Risk factors for breast cancer have been well characterized. Breast cancer is 100 times more frequent in women than in men.\
Factors associated with an increased exposure to estrogen have also been elucidated including early menarche, late menopause, later age\
at first pregnancy, or nulliparity. The use of hormone replacement therapy has been confirmed as a risk factor, although mostly limited to \
the combined use of estrogen and progesterone, as demonstrated in the WHI (2). Analysis showed that the risk of breast cancer among women using \
estrogen and progesterone was increased by 24% compared to placebo. A separate arm of the WHI randomized women with a prior hysterectomy to \
conjugated equine estrogen (CEE) versus placebo, and in that study, the use of CEE was not associated with an increased risk of breast cancer (3).\
Unlike hormone replacement therapy, there is no evidence that oral contraceptive (OCP) use increases risk. A large population-based case-control study \
examining the risk of breast cancer among women who previously used or were currently using OCPs included over 9,000 women aged 35 to 64 \
(half of whom had breast cancer) (4). The reported relative risk was 1.0 (95% CI, 0.8 to 1.3) among women currently using OCPs and 0.9 \
(95% CI, 0.8 to 1.0) among prior users. In addition, neither race nor family history was associated with a greater risk of breast cancer among OCP users."
words = txt
corpus = " ".join(words).lower()
sentences1 = sent_tokenize(corpus)
a = [" ".join([sentences1[i-1],j]) for i,j in enumerate(sentences1) if [item in List1] in word_tokenize(j)]
for i in a:
print i,'\n','\n'
foo()
The problem is that the python IDLE does not print anything. What could I have done wrong. What it does is run the code and I get this
>
>
Your question isn't very clear to me so please correct me if i'm getting this wrongly. Are you trying to match the list of keywords (in list1) against the text (in txt)? That is,
For each keyword in list1
Do a match against every sentences in txt.
Print the sentence if they matches?
Instead of writing a complicated regular expression to solve your problem I have broken it down into 2 parts.
First I break the whole lot of text into a list of sentences. Then write simple regular expression to go through every sentences. Trouble with this approach is that it is not very efficient but hey it solves your problem.
Hope this small chunk of code can help guide you to the real solution.
def foo():
List1 = ['risk','cancer','ocp','hormone','OCP',]
txt = "blah blah blah - truncated"
words = txt
matches = []
sentences = re.split(r'\.', txt)
keyword = List1[0]
pattern = keyword
re.compile(pattern)
for sentence in sentences:
if re.search(pattern, sentence):
matches.append(sentence)
print("Sentence matching the word (" + keyword + "):")
for match in matches:
print (match)
--------- Generate random number -----
from random import randint
List1 = ['risk','cancer','ocp','hormone','OCP',]
print(randint(0, len(List1) - 1)) # gives u random index - use index to access List1
I have a text file which has information, like so:
product/productId: B000GKXY4S
product/title: Crazy Shape Scissor Set
product/price: unknown
review/userId: A1QA985ULVCQOB
review/profileName: Carleen M. Amadio "Lady Dragonfly"
review/helpfulness: 2/2
review/score: 5.0
review/time: 1314057600
review/summary: Fun for adults too!
review/text: I really enjoy these scissors for my inspiration books that I am making (like collage, but in books) and using these different textures these give is just wonderful, makes a great statement with the pictures and sayings. Want more, perfect for any need you have even for gifts as well. Pretty cool!
product/productId: B000GKXY4S
product/title: Crazy Shape Scissor Set
product/price: unknown
review/userId: ALCX2ELNHLQA7
review/profileName: Barbara
review/helpfulness: 0/0
review/score: 5.0
review/time: 1328659200
review/summary: Making the cut!
review/text: Looked all over in art supply and other stores for "crazy cutting" scissors for my 4-year old grandson. These are exactly what I was looking for - fun, very well made, metal rather than plastic blades (so they actually do a good job of cutting paper), safe ("blunt") ends, etc. (These really are for age 4 and up, not younger.) Very high quality. Very pleased with the product.
I want to parse this into a dataframe with the productID, title, price.. as columns and the data as the rows. How can I do this in R?
A quick and dirty approach:
mytable <- read.table(text=mytxt, sep = ":")
mytable$id <- rep(1:2, each = 10)
res <- reshape(mytable, direction = "wide", timevar = "V1", idvar = "id")
There will be issues if there are other colons in the data. Also assumes that there is an equal number (10) of variables for each case. All
This question is in the context of twoway line with the by() option, but I think the bigger problem is how to identify a second (and all subsequent) event windows without a priori knowing every event window.
Below I generate some data with five countries over the 1990s and 2000s. In all countries an event occurs in 1995 and in Canada only the event repeats in 2005. I would like to plot outcome over the five years centered on each event in each country. If I do this using twoway line and by(), then Canada plots twice in the same plot window.
clear
set obs 100
generate year = 1990 + mod(_n, 20)
generate country = "United Kingdom" in 1/20
replace country = "United States" in 21/40
replace country = "Canada" in 41/60
replace country = "Australia" in 61/80
replace country = "New Zealand" in 81/100
generate event = (year == 1995) ///
| ((year == 2005) & (country == "Canada"))
generate time_to_event = 0 if (event == 1)
generate outcome = runiform()
encode country, generate(countryn)
xtset countryn year
forvalue i = 1/2 {
replace time_to_event = `i' if (l`i'.event == 1)
replace time_to_event = -`i' if (f`i'.event == 1)
}
twoway line outcome time_to_event, ///
by(country) name(orig, replace)
A manual solution adds an occurrence variable that numbers each event occurrence by country, then adds occurrence to the by() option.
generate occurrence = 1 if !missing(time_to_event)
replace occurrence = 2 if ///
(inrange(year, 2005 - 2, 2005 + 2) & (country == "Canada"))
twoway line outcome time_to_event, ///
by(country occurrence) name(attempt, replace)
This works great in the play data, but in my real data there are many more countries and many more events. I can manually code this occurrence variable, but that is tedious (and now I'm really curious if there's a tool or logic that works :) ).
Is there a logic to automate identifying windows? Or one that at least works with twoway line? Thanks!
You have generated a variable time_to_event which is -2 .. 2 in a window and missing otherwise. You can use tsspell from SSC, installed by
ssc inst tsspell
to label such windows. Windows are defined by spells or runs of observations all non-missing on that time_to_event:
tsspell, cond(time_to_event < .)
tsspell requires a prior tsset and generates three variables explained in its help. You can then renumber windows by using one of those variables _seq (sequence number within spell, numbered 1 up)
gen _spell2 = (_seq > 0) * sum(_seq == 1)
and then label spells distinctly by using country and the spell identifier for each spell from _spell, another variable produced by tsspell:
egen gspell = group(country _spell) if _spell2, label
My code assumes that windows are disjoint and cannot overlap, but that seems to be one of your assumptions too. Some technique for handling spells is given by http://www.stata-journal.com/sjpdf.html?articlenum=dm0029 That article does not mention tsspell, which in essence is an implementation of its principles. I started explaining the principles, but the article got long enough before I could explain the program. As the help of tsspell is quite detailed, I doubt that a sequel paper is needed, or at least that it will be written.
(LATER) This code also assumes that windows don't touch. Solving that problem suggests a more direct approach not involving tsspell at all:
bysort country (year) : gen w_id = (time_to_event < .) * sum(time_to_event == -2)
egen w_label = group(country w_id) if w_id, label