LINK1 = "https://en.wikipedia.org/wiki/List_of_Wimbledon_gentlemen%27s_singles_champions"library(htmltab)
B = htmltab(doc = LINK1,
which = '//*[@id="mw-content-text"]/div/table[4]' )names (B)## [1] "Year" "Country" "Champion"
## [4] "Country" "Runner-up" "Score in the final"str (B)## 'data.frame': 52 obs. of 6 variables:
## $ Year : chr "1968" "1969" "1970" "1971" ...
## $ Country : chr " AUS" " AUS" " AUS" " AUS" ...
## $ Champion : chr "Rod Laver" "Rod Laver" "John Newcombe" "John Newcombe" ...
## $ Country : chr " AUS" " AUS" " AUS" " USA" ...
## $ Runner-up : chr "Tony Roche" "John Newcombe" "Ken Rosewall" "Stan Smith" ...
## $ Score in the final: chr "6–3, 6–4, 6–2" "6–4, 5–7, 6–4, 6–4" "5–7, 6–3, 6–2, 3–6, 6–1" "6–3, 5–7, 2–6, 6–4, 6–4" ...#Recodificación
names(B)[6]="Score"B[,c(1)]=lapply(B[,c(1)], as.numeric) ## Warning in `[<-.data.frame`(`*tmp*`, , c(1), value = list(1968, 1969,
## 1970, : provided 52 variables to replace 1 variablesstr(B)## 'data.frame': 52 obs. of 6 variables:
## $ Year : num 1968 1968 1968 1968 1968 ...
## $ Country : chr " AUS" " AUS" " AUS" " AUS" ...
## $ Champion : chr "Rod Laver" "Rod Laver" "John Newcombe" "John Newcombe" ...
## $ Country : chr " AUS" " AUS" " AUS" " USA" ...
## $ Runner-up: chr "Tony Roche" "John Newcombe" "Ken Rosewall" "Stan Smith" ...
## $ Score : chr "6–3, 6–4, 6–2" "6–4, 5–7, 6–4, 6–4" "5–7, 6–3, 6–2, 3–6, 6–1" "6–3, 5–7, 2–6, 6–4, 6–4" ...table (B$Country)##
## SUI AUS CRO ESP FRG GBR GER NED SRB SWE TCH USA
## 8 6 1 2 3 2 1 1 5 7 1 15library(questionr)
library(magrittr)
B1=freq(B$Country,total = F,sort = 'dec',exclude = c(NA)) %>% data.frame()
B1=data.frame(variable=row.names(B1),B1,row.names = NULL)
B1## variable n X.
## 1 USA 15 28.8
## 2 SUI 8 15.4
## 3 SWE 7 13.5
## 4 AUS 6 11.5
## 5 SRB 5 9.6
## 6 FRG 3 5.8
## 7 ESP 2 3.8
## 8 GBR 2 3.8
## 9 CRO 1 1.9
## 10 GER 1 1.9
## 11 NED 1 1.9
## 12 TCH 1 1.9library(ggplot2)
base = ggplot(data=B1,aes(x=variable,y=n))
B2 = base + geom_bar(stat='identity')
B2 = B2 + scale_x_discrete(limits = B1$variable)
B2= B2 + labs(title="Países con mayor cantidad de ganadores del Wimbledon ",
subtitle = "De 1968 al 2019 ",
x ="País",
y = "Conteo",
caption = "Fuente: Wikipedia")
B2
library(qcc)## Package 'qcc' version 2.7## Type 'citation("qcc")' for citing this R package in publications.pareto.chart(table(B$Country),cumperc = c(0, 25,50,75,100))
##
## Pareto chart analysis for table(B$Country)
## Frequency Cum.Freq. Percentage Cum.Percent.
## USA 15.000000 15.000000 28.846154 28.846154
## SUI 8.000000 23.000000 15.384615 44.230769
## SWE 7.000000 30.000000 13.461538 57.692308
## AUS 6.000000 36.000000 11.538462 69.230769
## SRB 5.000000 41.000000 9.615385 78.846154
## FRG 3.000000 44.000000 5.769231 84.615385
## ESP 2.000000 46.000000 3.846154 88.461538
## GBR 2.000000 48.000000 3.846154 92.307692
## CRO 1.000000 49.000000 1.923077 94.230769
## GER 1.000000 50.000000 1.923077 96.153846
## NED 1.000000 51.000000 1.923077 98.076923
## TCH 1.000000 52.000000 1.923077 100.000000El 80% del total de ganadores está representado por los Estados Unidos, Suecia, Suiza, Australia y Serbia.
library (DescTools)
Mode(B$Country)## [1] " USA"La moda es Estados Unidos, por ende, el país que más veces ha ganado el certamen es Estados Unidos.
Herfindahl(table(B$Country))## [1] 0.1553254La moda no es significativa, Estados Unidos no posee una preponderancia cabal.
1/Herfindahl(table(B$Country))## [1] 6.438095Hay 6 grupos representativos.
LINK2="https://en.wikipedia.org/wiki/List_of_Wimbledon_ladies%27_singles_champions"C = htmltab(doc = LINK2,
which = '//*[@id="mw-content-text"]/div/table[4]' )names (C)## [1] "Year" "Country" "Champion"
## [4] "Country" "Runner-up" "Score in the final"str (C)## 'data.frame': 52 obs. of 6 variables:
## $ Year : chr "1968" "1969" "1970" "1971" ...
## $ Country : chr " USA" " GBR" " AUS" " AUS" ...
## $ Champion : chr "Billie Jean King" "Ann Jones" "Margaret Court" "Evonne Goolagong" ...
## $ Country : chr " AUS" " USA" " USA" " AUS" ...
## $ Runner-up : chr "Judy Tegart" "Billie Jean King" "Billie Jean King" "Margaret Court" ...
## $ Score in the final: chr "9–7, 7–5" "3–6, 6–3, 6–2" "14–12, 11–9" "6–4, 6–1" ...Sólo se empleará la variable “Country” del ganador del torneo, por ende, las variables correspondientes de quien perdió no son necesarias.
#Recodificación
C = C [,c (1:3,6)]
C[,c(1)]=lapply(C[,c(1)], as.numeric) ## Warning in `[<-.data.frame`(`*tmp*`, , c(1), value = list(1968, 1969,
## 1970, : provided 52 variables to replace 1 variablesstr (C)## 'data.frame': 52 obs. of 4 variables:
## $ Year : num 1968 1968 1968 1968 1968 ...
## $ Country : chr " USA" " GBR" " AUS" " AUS" ...
## $ Champion : chr "Billie Jean King" "Ann Jones" "Margaret Court" "Evonne Goolagong" ...
## $ Score in the final: chr "9–7, 7–5" "3–6, 6–3, 6–2" "14–12, 11–9" "6–4, 6–1" ...table (C$Country)##
## SUI AUS CZE ESP FRA FRG GBR GER ROU RUS USA
## 1 3 3 2 2 2 2 6 1 1 29C1=freq(C$Country, total = F,sort = 'dec',exclude = c(NA)) %>% data.frame()
C1=data.frame(variable=row.names(C1),C1,row.names = NULL)
C1## variable n X.
## 1 USA 29 55.8
## 2 GER 6 11.5
## 3 AUS 3 5.8
## 4 CZE 3 5.8
## 5 ESP 2 3.8
## 6 FRA 2 3.8
## 7 FRG 2 3.8
## 8 GBR 2 3.8
## 9 SUI 1 1.9
## 10 ROU 1 1.9
## 11 RUS 1 1.9base = ggplot(data=C1,aes(x=variable,y=n))
C2 = base + geom_bar(stat='identity')
C2= C2 + labs(title="Ganadoras del Wimbledon por país",
subtitle = "Desde al año 1968 a la actualidad", x = "Países" ,
y = "Cantidad",
caption =" Fuente: Wikipedia")
C2 + theme_classic() +
theme(plot.title = element_text(hjust = 0.5,size=20),
plot.subtitle = element_text(hjust = 1),
plot.caption = element_text(hjust = 1),
axis.text.x = element_text(angle=45,hjust = 1))
pareto.chart(table(C$Country),cumperc = c(0,50,80,100))
##
## Pareto chart analysis for table(C$Country)
## Frequency Cum.Freq. Percentage Cum.Percent.
## USA 29.000000 29.000000 55.769231 55.769231
## GER 6.000000 35.000000 11.538462 67.307692
## AUS 3.000000 38.000000 5.769231 73.076923
## CZE 3.000000 41.000000 5.769231 78.846154
## ESP 2.000000 43.000000 3.846154 82.692308
## FRA 2.000000 45.000000 3.846154 86.538462
## FRG 2.000000 47.000000 3.846154 90.384615
## GBR 2.000000 49.000000 3.846154 94.230769
## SUI 1.000000 50.000000 1.923077 96.153846
## ROU 1.000000 51.000000 1.923077 98.076923
## RUS 1.000000 52.000000 1.923077 100.000000Los Estados Unidos poseen más del 50% de ganadoras. El 80% está representado por Estados Unidos, Alemania, Australia, República Checa y España.
Mode(C$Country)## [1] " USA"La moda es Estados Unidos, el país más ganador del Wimbledon es Estados Unidos.
Herfindahl(table(C$Country))## [1] 0.3380178La moda es significativa, por ende, se puede interpretar que los Estados Unidos incentivo mucho este deporte.
1/sum(prop.table(table(C$Country))**2)## [1] 2.958425Existen dos grupos representativos, lo que implica que estos países se preocupan por el deporte.