Alejandro Carrera's questions

Suppose we have a string of last names like the following:

nom <- c("Perez Conchito", "Juanin Juanharry", "Von Bola")

I have designed a small function to extract each part of the string and then gather them into a data.frame with two columns.

extract_apellidos <- function(x) {
    first_split <- strsplit(x, " ")
    split_un <- unlist(first_split)
    primer_ap <- split_un[c(TRUE, FALSE)] #Nos da el primer apellido
    segundo_ap <- split_un[c(FALSE,TRUE)] #Nos da el segundo apellido
    data.frame(Primer_ap=primer_ap, Segundo_ap=segundo_ap)
}

extract_apellidos(nom)


  Primer_ap Segundo_ap
1     Perez   Conchito
2    Juanin  Juanharry
3       Von       Bola

As you can see, it works correctly. However, I would like to know if it is possible to optimize it using regular functions since I suspect that this would allow me to reduce the number of steps used. I thank you in advance for any guidance on this.

I'm trying to create a vector of dates for, say, the period from January 1, 2018 to October 31, 2019 using base R. What I can think of is to first create that vector using the seqand functions as.Dateas follows:

fecha <- data.frame(fecha=seq(from = as.Date("2018-01-01", format="%Y-%m-%d"), 
    to = as.Date("2019-10-31", format="%Y-%m-%d"), by = 1))

Which returns the following:

str(fecha)
## $ fecha: Date, format: "2018-01-01" "2018-01-02" 

Now, we commonly use the day-month-year format, and, to have that format, we use the function format.

fecha[,"fecha"] <- format(fecha[,"fecha"], "%d-%m-%Y")
str(fecha)
'data.frame':   669 obs. of  1 variable:
 $ fecha: chr  "01-01-2018" "02-01-2018" "03-01-2018" "04-01-2018" ...

The format is the desired one but the dates are now of character type. If I try to coerce using again as.DateI pretty much go back to the beginning:

fecha[,"fecha"] <- as.Date(fecha[,"fecha"], format="%d-%m-%Y")
str(fecha)
'data.frame':   669 obs. of  1 variable:
 $ fecha: Date, format: "2018-01-01" "2018-01-02" ...

Any solution to this problem? I am interested in having one based on R and I would appreciate any additional solution (I understand that the libraries allow zooyou lubridateto work with dates, although I honestly don't know their main functions)

Thanks in advance for any guidance.

I have a database similar to this:

library(stringr) #Para la función str_extract

ejemplo <- data.frame(columna= c("{ em , 680/3 }", "no 659-11", "no funciona la 2507", 
    "a-4-2 no funciona", "p-8 con presencia de arena", " no 2-5c12s si sirve",
    "mecanica no 1ty -22s brinca"))

I am interested in extracting the part of each string that is:

• Number
• letter plus number
• number plus letter

and, in general, strings referring to "identifiers".

Using a simple regular expression, I can extract

ejemplo$col2 <- ifelse(grepl("[0-9]+", ejemplo$columna), str_extract(ejemplo$columna, "[0-9]+"), NA)
ejemplo

                     columna col2
1              { em , 680/3 }  680
2                   no 659-11  659
3         no funciona la 2507 2507
4           a-4-2 no funciona    4
5  p-8 con presencia de arena    8
6         no 2-5c12s si sirve    2
7 mecanica no 1ty -22s brinca    1

which only has a good result for the third string. Honestly I feel lost and I don't know if there is a way to do everything I want in one go or I will have to build several regular expressions to achieve it. I greatly appreciate any guidance on this.

The desired output would be:

  deseado
1   680/3
2  659-11
3    2507
4   a-4-2
5     p-8
6 2-5c12s
7 1ty-22s

I have a column with text similar to the following:

cadenas <- c("el metro", "dos metrobuses", "pasó en el metro norte", "metro 12", "no")

I'm interested in only getting the elements that strictly have the word "meter", that is, elements 1, 3 and 4.

grepl("metro", cadenas) 

Returns TRUE for the first 4 and FALSE for the last. I have tried many combinations to get only the ones that interest me without any success.

cadenas[grepl("metro [[:alnum:]]", cadenas)] #Regresa los elementos 3 y 4 (resultado más cercano al deseado)
cadenas[grepl("\\ metro [[:alnum:]]", cadenas)] #Regresa el elemento 3
cadenas[grepl("\\ metro", cadenas)] #Regresa los elementos 1,2 y 3

I have read many places without finding an answer that satisfies my requirements so any guidance would be greatly appreciated.

On many occasions, when we work with databases, we find various encodings for unregistered or unreported data. For example, in can be zeros, -999, -99, among others, which, for data processing purposes, we can convert to NA.

Thinking about it, I made a small function that looks for unregistered values ​​that can be changed to NA:

'is.na.m<-' <- function(x, value, ...) {
    x[c(value)] = NA
    x
}

We create a test vector:

x <- c(1:3,2:5,1:10, -99, -999, -98)
is.na.m(x) = x%in%c(-99, -999, -98)
x

Salida:
[1]  1  2  3  2  3  4  5  1  2  3  4  5  6  7  8  9 10 NA NA NA

Now, I've tried to get this to work on a dataframe, but to no avail:

b <-data.frame(animal=c("perro", "gato", -999), num=c(1,-98,3))
is.na.m(b) = b[b%in%c(-98, -999)]
b

As can be seen, there are no changes in the dataframe:

  animal num
1  perro   1
2   gato -98
3   -999   3

** It should be noted that is.na.m(b) = b%in%c(-98, -999)it did not work either

I tried to use the indexing function, but it didn't work either:

is.na.m(b[,1:ncol(b)]) = b[,1:ncol(b)%in%c(-98, -999)]
b

Now, when I try to use lapply, it gives me an error:

b <- unlist(lapply(b, is.na.m(b))
Error in is.na.m(b) : no se pudo encontrar la función "is.na.m"

The question is: What are the adjustments that I must make in the function so that it operates correctly in all the columns of a dataframe?

I thank you in advance for any guidance.

I'm taking a course on data visualization in Python and right now we're looking at some basic stuff about HTML and Javascript.

The following code makes a simple HTML box using the command %%hmtlfrom a Jupyter notebook.

## Esto se ejecuta en una celda

%%html

<style>
    #boton
    {
        width: 60px;
        height: 50px;
        background-color: grey;
        color: white;
        text-align: center;
    }
</style>
<div id="boton"></div>

Now, to create events on the box, we use %%javascript. In order for the function to register the double click and change the color of the box, I followed the model that comes in this tutorial:

https://www.w3schools.com/js/js_htmldom_eventlistener.asp

The code would look like this:

##Esto se ejecuta en otra celda
    %%javascript

    let boton = document.getElementById("boton");
    boton.addEventListener("click", myf1);
    boton.addEventListener("click", myf2);
    boton.addEventListener("mouseover", function(){boton.textContent="¡Hola!"});
    boton.addEventListener("mouseout", function(){boton.textContent="No te vayas"});


    function myf1() {
      document.getElementById("boton").style.backgroundColor= "red";
    }

    function myf2() {
      document.getElementById("boton").style.backgroundColor = "blue";
    }

Now when I return to the button, the color immediately changes to blue, i.e. it doesn't log the "red" then "blue" events on click. Considering that I am a beginner, I appreciate any guidance and also your understanding and patience.

To make my question clearer, I put an example in code, Rwhich is the language in which I have a little more experience. Suppose I have the simple variable "a" that is equal to 5. First of all I want to check if it is numeric using is.numeric:

a <- 5
is.numeric(a)
[1] TRUE

Now, to check if it is not numeric, we use a ! before the method and we get the following:

!is.numeric(a)
[1] FALSE

My question is if it is possible to do something like this in Python, "negating" a method like isinstance() that follows the structure:

isinstance(variable, tipo)

For example, if we apply negation to isinstance, I get:

a= 5
!isinstance(a, int)

"isinstance" no se reconoce como un comando interno o externo,
programa o archivo por lotes ejecutable.

I didn't find any useful information, that's why I pose my question. If there is material that I have overlooked, I would also greatly appreciate if you would let me know.

I'm going over some Python basics and doing exercises on it. One of them is to perform a Fibonacci sequence, which I solved using a whileand a pair of operators +=. My question is if this code can be made simpler using loops.

a=1
b=1

print(a)
print(b)
while True:
    a+=b
    b+=a
    if b>1000:
        break
    print(a)
    print(b)


Salida:

1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987

Thanks in advance :)

I was reviewing some things about matrices and the Mahalanobis distance and it occurred to me to make a small function that ranks the observations of each column of a matrix. Below the code:

test <- matrix(c(78.17,70.25,75.33, 86.08,54.97, 43.63,18.04,
0.3,1.4,0.5,1.5,0.7,0.2,0.1,3,5,5,8,9,10,2), ncol=3)
test

rank_columns <- function (x) {
    y <- matrix(ncol=ncol(x), nrow=nrow(x))    
    for (j in 1:ncol(x)) {
    y[,j] <- rank(x[,j])
        }
        return(y)
    }

rank_columns (test)

The function returns an array with the original dimensions of the input array and the ranked observations:

rank_columns(test)
     [,1] [,2] [,3]
[1,]    6    3  2.0
[2,]    4    6  3.5
[3,]    5    4  3.5
[4,]    7    7  5.0
[5,]    3    5  6.0
[6,]    2    2  7.0
[7,]    1    1  1.0

As many of you know, I'm not very good at using the family apply, so I was wondering if there was a way to vectorize the function to optimize its performance when dealing with larger matrices. Beforehand thank you very much.

Before presenting my problem, I share the link where you can download the document I work with:

https://drive.google.com/file/d/178s_tfbqbXmnxsknxF8DP154_N1DYjgf/view

Here is the code I am using: (Don't forget to create your own route before running it)

library(rJava)
library(tm)
library(qdap)
library(tidyverse)
library(pdftools)   
library(stringr)
library(tidytext)
library(stringi)
library(wordcloud)

stop_es <- c(stopwords("es"))   #Vector de palabras "innecesarias" 

cce <- pdf_text("ruta_del_archivo")     #Para leer el archivo
corpus <- Corpus(VectorSource(cce))  #Crear corpus

#Limpieza y preprocesamiento
CCE  <- tm_map(corpus, tolower) %>% 
tm_map(stripWhitespace) %>%
tm_map(removePunctuation) %>% 
tm_map (removeWords, stop_es) %>%
stri_trans_general("Latin-ASCII")   #Con esto removemos acentos en español

##Crear nuevamente el corpus. Por alguna razón que no entiendo, stri_trans_general crea un archivo que no puede leerse si no se transforma nuevamente en corpus 

CCEGTO <- Corpus(VectorSource(CCE))

After the previous steps, we proceed to create the table of most frequent terms

ft <- freq_terms(CCEGTO, 50, stopwords=stop_es)
ft

The output shows words that seem to be incomplete or don't make sense: (I removed some to focus my attention on the problem)

   WORD        FREQ
2  ca           105  ##¿?
3  guanajuato    94
5  vo            86
6  ufb           75   ##¿¿??
9  va            69
10 propuestas    68
11 nivel         64
12 par           58    #Esto podría ser la raíz de "parte", "participacion" o alguna por el estilo
27 ins           42    #Esto podría ser para "instituto", "institucion" o algo parecido 
28 n             42    #Una sola letra como término frecuente... ¿?
30 vos           41
33 numero        40
34 vas           40
35 l             39
38 d             37
39 s             37
42 poli          35  #Esta podría ser para "policia", "politica", "politicas"
43 vidad         35  #Esta podría ser para "vida" o"actividad"
44 cas           34
45 r             34   #Un caracter aislado
46 cipacion      33   #Esta podría ser la primera parte de la palabra "participacion" 
47 i             33   #Otra letra aislada...

My fundamental question is if I am doing something wrong with the pre-processing of the text or if it is the structure of the pdf file itself that is not allowing me to do a good job of parsing.

Any comments and suggestions are greatly appreciated.

I try to apply a simple normalization function to the numeric variables in the R database iristhrough an forand using lapplyin order to obtain a new database containing only the normalized variables:

data(iris)

normal <- function (x) {
num <- x - min(x)
den <- max(x) - min(x)
return (num/den)
}

iris_n <- data.frame()

for (i in 1:length(iris)){
if (is.numeric(iris[,i])) {
}
iris_n[,i] <- as.data.frame(lapply(iris[,i], normal))
}

Error in Summary.factor(1L, na.rm = FALSE) : 
  ‘min’ not meaningful for factors
Además: There were 50 or more warnings (use warnings() to see the first 50)

iris_n

[1] NaN.   NaN..1 NaN..2 NaN..3
<0 rows> (or 0-length row.names)

Now, I try to lapply directly with:

iris_n <- as.data.frame(lapply(iris, function(x) {if (is.numeric(x))  normal(x)}))

Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE,  : 
  arguments imply differing number of rows: 150, 0

No matter how hard I turn it, I can't find the error. Any guidance will be greatly appreciated (Note: I am interested in solving this specific problem, I know there are other types of solutions to achieve what I need)

Preliminary:

First of all, I apologize for not uploading my full code as it's a bit long and I don't know how to upload a file here. If anyone knows please guide me so I can update my question.

I downloaded from this website an excel file:

datosabiertos.segob.gob.mx/DatosAbiertos/sesnsp_Incidencia_delictiva_Fuero_Federal/IDEFF

I saved it as csv and after multiple manipulations to the database I have this:

> head(d1217full)
   año     mes comercio contra la salud narcomenudeo otros otros_lfcd otros_lgs transporte posesion produccion trafico total
1 2012   enero     0.16               0         5.87  0.13       0.02      0.00       0.02     4.34       0.07       0 10.61
2 2012 febrero     0.04               0         5.30  0.13       0.02      0.00       0.00     4.68       0.02       0 10.19
3 2012   marzo     0.07               0         4.81  0.07       0.05      0.00       0.00     6.12       0.07       0 11.21
4 2012   abril     0.40               0         4.48  0.13       0.04      0.00       0.00     4.85       0.00       0  9.90
5 2012    mayo     0.36               0         5.91  0.36       0.07      0.00       0.02     1.95       0.05       0  8.73
6 2012   junio     0.26               0         5.30  0.15       0.22      1.48       0.02     2.17       0.00       0  9.59

When applying strwe have:

> str(d1217full)
'data.frame':   72 obs. of  13 variables:
 $ año            : Factor w/ 6 levels "2012","2013",..: 1 1 1 1 1 1 1 1 1 1 ...
 $ mes            : Factor w/ 12 levels "enero","febrero",..: 1 2 3 4 5 6 7 8 9 10 ...
 $ comercio       : num  0.16 0.04 0.07 0.4 0.36 0.26 0.09 0.04 0.04 0.05 ...
 $ contra la salud: num  0 0 0 0 0 0 0 0 0 0.02 ...
 $ narcomenudeo   : num  5.87 5.3 4.81 4.48 5.91 5.3 4.27 6.6 6.29 9.84 ...
 $ otros          : num  0.13 0.13 0.07 0.13 0.36 0.15 0.13 0.09 0.02 0.09 ...
 $ otros_lfcd     : num  0.02 0.02 0.05 0.04 0.07 0.22 0.11 0.11 0.13 0.11 ...
 $ otros_lgs      : num  0 0 0 0 0 1.48 4.12 0.2 0.46 0.51 ...
 $ transporte     : num  0.02 0 0 0 0.02 0.02 0 0.02 0.02 0.02 ...
 $ posesion       : num  4.34 4.68 6.12 4.85 1.95 2.17 2.11 1.91 1.71 1.73 ...
 $ produccion     : num  0.07 0.02 0.07 0 0.05 0 0.02 0 0 0 ...
 $ trafico        : num  0 0 0 0 0 0 0 0 0 0 ...
 $ total          : num  10.61 10.19 11.21 9.9 8.73 ...

What I want is to be able to make a graph that has the following characteristics:

1. On the x-axis the year and month from 2012/01 to 2017/12 are recorded.
2. On the y-axis, with different lines, the data of the trade variables are shown up to total, that is, a total of 11 lines, emphasizing "total"

Since my time variables are factors, ggplotit doesn't allow me to graph properly. For example,

ggplot(d1217full, aes(año, total)) +
geom_line()

It results in this plot, which of course is wrong.

The foregoing, without considering that it would still be necessary to include the rest of the variables in the plot.

I thought about creating a time vector from scratch, bringing it into my database, and then making a graph following something like:

añosgto <- c(seq(as.Date("2012/1/1"), by = "month", length.out = 72))
añosgto <- as.character(añosgto)
temp  <- strsplit(añosgto, "-")
temp <- matrix(unlist(temp), ncol=3, byrow=T)
temp <- as.data.frame(temp)
temp <- mutate(temp, año_def=V1, mes_def=V2)
temp <- temp[,-(1:3)]
head(temp)

D1217 <- cbind(d1217full, temp)
D1217 <- select(D1217, -(1:2))

ggplot(D1217, aes(str_c(año_def, "/", str_pad(mes_def, 2, pad = 0)),
  total, group=total)) +
  geom_line() +
   scale_x_discrete(breaks = c("2012/12", "2013/12", "2014/12", "2015/12", "2016/12", "2017/12"))

Which results in this:

As you can see, I've tried graphing without success so any guidance would be greatly appreciated.

UPDATE:

Create a repository on github where the database and code are located.

https://github.com/LuisAlejandroCG/project_gto

First of all, I want to comment that I did an exhaustive search before asking my question without obtaining satisfactory results. The database names I want to use are quite long and confusing and therefore I want to change their name but using their position. Some of the variable names are:

 [1] "Fecha"                                                                                    
 [2] "Delegación"                                                                               
 [3] "Clave.INEGI.AGEE"                                                                         
 [4] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_.Producción"                           
 [5] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Transporte"                            
 [6] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Tráfico"                               
 [7] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Comercio"                              
 [8] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Suministro"                            
 [9] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Posesión"                              
[10] "Código.Penal.Federal..CPF._Delitos.contra.la.salud_Otros"                                 

It's easy to change the name with rename using:

rename(nuevo_nombre = viejo_nombre)

when it comes to not so long names. However, this is not the case. Now I found solutions like:

rename(!!produccion := names[4], !!transporte:= names[5],
!!trafico :=names[6], !!comercio:= names[7], !!suministro := names[8],
!!posesion := names[9], !!otros:= names[10])

Error in quos(...) : objeto 'produccion' no encontrado

either:

rename("produccion" = names[4], "transporte"= names[5],
"trafico" =names[6], "comercio"= names[7], "suministro" = names[8],
"posesion" = names[9], "otros"= names[10])

Error: Expressions are currently not supported in `rename()`

The second obviously means that the format was valid in some earlier version of dplyr but not now. My question is: Is it no longer possible to rename a variable by position using renamede dplyr? What alternatives are there, particularly if I want to use the %>%given operator as it provides an accurate reading of how the database is being manipulated?

I appreciate any feedback and guidance.

Here is a very small view of the variables of a database that I am using for an exercise on classification.

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 41188 entries, 0 to 41187
Data columns (total 21 columns):
age               41188 non-null int64
job               41188 non-null object
marital           41188 non-null object

dtypes: float64(5), int64(6), object(10)
memory usage: 6.6+ MB

I want to normalize the variable ageand for that create a simple function:

def normalize(X):
    NORM = []
    for i in X:
        norm= (i-min(X))/(max(X)-min(X))
        NORM.append(norm)
    return NORM

The problem is that when I try to apply with mapor with applyto create a new variable I get the same error:

log["age_n"]=  log["age"].apply(normalize)
TypeError: 'int' object is not iterable

log["age_n"]=  log["age"].map(normalize)
TypeError: 'int' object is not iterable

Any guidance on this error would be greatly appreciated.

The "car" package in R has an interesting function called outlierTest and I want to analyze its source code to understand how it works. I tried the following and got an error:

getMethod("outlierTest")

    Error in getMethod("outlierTest") : 
     no generic function found for 'outlierTest'

This seems to get me a response but it redirects me to a page with no source code displayed:

methods(outlierTest)
    [1] outlierTest.lm*
    see '?methods' for accessing help and source code

Finally, when typing only "outlierTest" in console I get no response:

outlierTest
    function (model, ...) 
    {
          UseMethod("outlierTest")
    }
<bytecode: 0x000000000cf4d3d0>
<environment: namespace:car>

I appreciate any guidance on this.

A little algorithm that converts numbers to a character string and then adds it is:

total= 0

cadena= "1.2,1.5,1.7"

for i in cadena.split(","):
    i= float(i)
    total+= i
print(total)

Now, suppose that the numbers come individually in quotes like this:

cadena= "1.2","1.5","1.7"

What is the command that I must use so that each number remains as a float and then add it using the same for? Using the initial code for the last string I get the following error:

AttributeError: 'tuple' object has no attribute 'split'

Any guidance will be greatly appreciated.

Suppose I have the following Python dictionary:

 letters= {0: 'a',
 1: 'b',
 2: 'c',
 3: 'd',
 4: 'e',
 5: 'f',
 6: 'g',
 7: 'h',
 8: 'i',
 9: 'j',
 10: 'k',
 11: 'l',
 12: 'm',
 13: 'n',
 14: 'o',
 15: 'p',
 16: 'q',
 17: 'r',
 18: 's',
 19: 't',
 20: 'u',
 21: 'v',
 22: 'w',
 23: 'x',
 24: 'y',
 25: 'z',
 26: ' '}

And I change the order of the keys and values ​​with the following dict comprehension:

encoding = {y:x for x,y in letters.items()}

The next thing I want to do is that, for each letter (key), its value is increased by 3 using the following:

ek= 3

That way, we would have:

'a': 3
'b': 4

and so on.

How can I achieve that without having to do it manually? I would like to put some idea but these two approaches were not successful:

encoding= {sum(ek) for x in encoding.values()}
TypeError: 'int' object is not iterable


encoding.update(sum(ek) in encoding.values())
TypeError: 'int' object is not iterable

Any help will be greatly appreciated. Cheers!

I am trying to create a function that every time it is called, it will generate a list of 100 random rolls of a die using the random library.

import random

def dice():
    dado = []
    for i in range(100):
        dado.append(random.choice(range(1,7)))
        return dado

dice()

When I run the function, it only returns one value. On the other hand, when I do this:

def dice():
    dado = []
    for i in range(100):
        dado.append(random.choice(range(1,7)))
        print(dado)

It returns 100 lists. The first contains one number, the next two, and so on until the list 100 with 100 numbers is printed. As I pointed out, what I want is for a single list of 100 random numbers to be printed.

I want to point out that I am also interested in the fact that the resulting list can be plotted through a histogram, for example.

Any help or guidance is greatly appreciated. Thank you!

I was doing an exercise that consisted of using a list comprehension (LC) to add the odd numbers between 0 and 10. To do this, it first occurred to me to use the for loop, analyze its structure and then pass it to an LC. The problem is that when using this for loop:

for i in range(10):
    if  i%2 != 0:
        print(sum([i]))

I didn't get the desired result, just a list of numbers. Instead, when using LC

sum([i for i in range(10) if i%2 !=0])

the sum was correct. Could someone guide me on the error? I'm pretty new to Python and I can't find any reference to help me solve this issue. Thanks in advance :)

I am trying to create a function, let's say "generic" to add rows in a database.

My idea is that the function "recognizes" the numeric elements in the database and adds by rows.

Suppose I have the data of the following image:

Screenshot:

The function would take from the second column and add 2+3+3+6+4+2+0 and so on.

The code I have written so far is as follows:

suma_renglones < - function (x) {
  y = integer(nrow(x))
  for (i in 1: nrow(x)) {
    for (j in 1: ncol(x)) {
      if (is.numeric(x[i, j])) {
        y[i] = y[i] + (x[i, j])
      }
    }
    y[i]
  }
}

However, when applying the function, the column designated to store the result of the sum shows only NULL results. Any help or guidance will be greatly appreciated.

PS: I know there are already solutions about it elsewhere as shown in this example , however my intention is to make a "own" function.