DF作为矩阵的总结
Summary of DF as matrix
我已经搜索了很长时间，但找不到真正简单的方法。
我有一个仅由数值组成的 df，我想从我的 df 中创建一个汇总矩阵。
DF
V1 V2 V3 V4 V5 ...
x1 y1 z1 1 c1
x2 NA z2 0 c2
x3 y3 z3 1 NA
...
V4本来是一个TRUE/FALSE变量转化为数值变量，应该可以正常使用。
我想获得以下内容：
N Mean SD Min 1st Median 3rd Max
V1
V2
V3
V4
V5
...
根据 N、平均值、SD、最小值、第 1、中值、第 3、最大值的值。
我试过简单的
as.data.frame(summary(DF))
我试过 stargazer 但由于某种原因不起作用（我猜是因为我有二进制变量）
stargazer(DF, type= "html", nobs = TRUE, type="html", mean.sd = TRUE, median = TRUE, iqr = TRUE,
+ digits=2, align=T)
我读了一些关于 qwraps2_summary_table 的文章。但是它们似乎都给出了与我正在寻找的 "design" 不同的 table。
我知道我也可以 运行 像这样的循环：
for(i in (1:length(DF)){
sum$N<-(????)
sum$Mean<-mean(DF[i])
....}
但这不是最好的解决方案。
有小费吗？谢谢！
这是我数据集的一部分
structure(list(Year = c(2011, 2012, 2013, 2014, 2015, 2016, 2017,
2018, 2018, 2011), Occurences = c(9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 2L, 9L), Balance = c(-1.14, 1.05, -1.06, 1.01, 1.01, 1.01,
-1.09, -1, -1.04, -1.03), Withdrawal = c(43200, 41080, 43400,
43183, 42600, 42100, 45900, 46000, 3892008, 48374), Verification_SA = c(NA,
NA, NA, NA, 1, 1, NA, 1, 1, NA), Classification_num = c(NA, NA,
NA, NA, 3, 2, NA, 4, 4, NA), Interaction_Verification_Classification = c(NA,
NA, NA, NA, 3, 2, NA, 4, 4, NA), KnowledgeSources = c(1, 1, 1,
0, 1, 1, 1, 1, 1, 0), KnowledgeDischarge = c(0, 0, 0, 0, 0, 1,
1, 1, 1, 0), Scarcity_watershed = c(NA_real_, NA_real_, NA_real_,
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_
), Scarcity_country = c(NA, NA, NA, NA, NA, NA, NA, NA, 3.35,
NA), Knowledge_Watershed = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Knowledge_Facilities = c(0,
0, 0, 0, 0, 0, 0, 0, 1, 1), Importance_num = c(NA, NA, NA, 3,
3, 3, 3, 3, 5, NA), DetrimentalImpacts_num = c(0, 0, 1, 0, 0,
0, 0, 0, 0, 0), Responsibility_num = c(1, 1, 1, 2, 2, 2, 2, 3,
3, 1)), row.names = c(NA, -10L), class = c("tbl_df", "tbl", "data.frame"
))
如果以后有人发现这个问题，请根据@camille 的建议检查 。这是一种简单的方法，尽管它没有提供 NA 的数量。
library(psych)
my_summary <- do.call(rbind,lapply(DF,psych::describe,quant=c(0.25,0.75)))
my_summary
# vars n mean sd median trimmed mad min max range skew kurtosis se Q0.25 Q0.75
#Year 1 10 2014.50 2.72 2014.50 2014.50 3.71 2011.00 2018.00 7.00 0.00 -1.74 0.86 2012.25 2016.75
#Occurences 1 10 8.30 2.21 9.00 9.00 0.00 2.00 9.00 7.00 -2.28 3.57 0.70 9.00 9.00
#Balance 1 10 -0.23 1.07 -1.02 -0.27 0.15 -1.14 1.05 2.19 0.35 -2.05 0.34 -1.06 1.01
#Withdrawal 1 10 428784.50 1216854.64 43300.00 44344.62 2535.25 41080.00 3892008.00 3850928.00 2.28 3.57 384803.22 42745.75 45975.00
#Verification_SA 1 4 1.00 0.00 1.00 1.00 0.00 1.00 1.00 0.00 NaN NaN 0.00 1.00 1.00
#Classification_num 1 4 3.25 0.96 3.50 3.25 0.74 2.00 4.00 2.00 -0.32 -2.08 0.48 2.75 4.00
...
这是我的原始解决方案，使用 data.table。
library(data.table)
my_summary <- rbindlist(lapply(DF, function(x){
as.data.frame(t(c(
summary(x),
SD = sd(x,na.rm=TRUE),
N = sum(!is.na(x)))))
})
, fill = TRUE, use.names = TRUE,idcol="Variable")
my_summary
# Variable Min. 1st Qu. Median Mean 3rd Qu. Max. SD N NA's
# 1: Year 2011.00 2012.250 2014.500 2014.500000 2016.75 2018.00 2.718251e+00 10 NA
# 2: Occurences 2.00 9.000 9.000 8.300000 9.00 9.00 2.213594e+00 10 NA
# 3: Balance -1.14 -1.055 -1.015 -0.228000 1.01 1.05 1.074800e+00 10 NA
# 4: Withdrawal 41080.00 42745.750 43300.000 428784.500000 45975.00 3892008.00 1.216855e+06 10 NA
# 5: Verification_SA 1.00 1.000 1.000 1.000000 1.00 1.00 0.000000e+00 4 6
# 6: Classification_num 2.00 2.750 3.500 3.250000 4.00 4.00 9.574271e-01 4 6
我们可以使用map循环遍历DF，获取summary统计数据，将其转换为data.frame，创建。 tibble 中的 'SD' 和 'N' 列以创建带有后缀的单个 data.frame 输出（_dfr in map）
library(purrr)
library(dplyr)
map_dfr(DF, ~ tibble(SD = sd(.x, na.rm = TRUE),
N = sum(!is.na(.x)), as.data.frame.list(base::summary(.x))),
.id = 'Variable')
# A tibble: 16 x 10
# SD N Min. X1st.Qu. Median Mean X3rd.Qu. Max. NA.s Variable
# * <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>
# 1 2.72e+0 10 2011 2012. 2014. 2.01e+3 2017. 2.02e3 NA Year
# 2 2.21e+0 10 2 9 9 8.30e+0 9 9.00e0 NA Occurences
# 3 1.07e+0 10 -1.14 -1.06 -1.02 -2.28e-1 1.01 1.05e0 NA Balance
# 4 1.22e+6 10 41080 42746. 43300 4.29e+5 45975 3.89e6 NA Withdrawal
# 5 0. 4 1 1 1 1.00e+0 1 1.00e0 6 Verification_SA
# 6 9.57e-1 4 2 2.75 3.5 3.25e+0 4 4.00e0 6 Classification_num
# 7 9.57e-1 4 2 2.75 3.5 3.25e+0 4 4.00e0 6 Interaction_Verificatio…
# 8 4.22e-1 10 0 1 1 8.00e-1 1 1.00e0 NA KnowledgeSources
# 9 5.16e-1 10 0 0 0 4.00e-1 1 1.00e0 NA KnowledgeDischarge
#10 NA 0 NA NA NA NaN NA NA 10 Scarcity_watershed
#11 NA 1 3.35 3.35 3.35 3.35e+0 3.35 3.35e0 9 Scarcity_country
#12 0. 10 0 0 0 0. 0 0. NA Knowledge_Watershed
#13 4.22e-1 10 0 0 0 2.00e-1 0 1.00e0 NA Knowledge_Facilities
#14 8.16e-1 6 3 3 3 3.33e+0 3 5.00e0 4 Importance_num
#15 3.16e-1 10 0 0 0 1.00e-1 0 1.00e0 NA DetrimentalImpacts_num
#16 7.89e-1 10 1 1 2 1.80e+0 2 3.00e0 NA Responsibility_num
要以 Ian Campbell 的回答为基础，在需要时不要害怕构建汇总函数。
summaryfn <- function(x){
c(min(x),
quantile(x,0.25,na.rm=TRUE),
quantile(x,0.5,na.rm=TRUE),
mean(x,na.rm=TRUE),
sd(x, na.rm=TRUE),
quantile(x,0.75,na.rm=TRUE),
max(x,na.rm=TRUE),
sum(is.na(x)))
}
res <- do.call(rbind,lapply(df,summaryfn))
colnames(res) <- c("Min","Q1","Med","Mean","Sd","Q3","Max","NAs")
## > res
## Min Q1 Med Mean Sd Q3 Max NAs
## Year 2011.00 2012.250 2014.500 2014.500000 2.718251e+00 2016.75 2018.00 0
## Occurences 2.00 9.000 9.000 8.300000 2.213594e+00 9.00 9.00 0
## Balance -1.14 -1.055 -1.015 -0.228000 1.074800e+00 1.01 1.05 0
## Withdrawal 41080.00 42745.750 43300.000 428784.500000 1.216855e+06 45975.00 3892008.00 0
## Verification_SA NA 1.000 1.000 1.000000 0.000000e+00 1.00 1.00 6
## Classification_num NA 2.750 3.500 3.250000 9.574271e-01 4.00 4.00 6
## Interaction_Verification_Classification NA 2.750 3.500 3.250000 9.574271e-01 4.00 4.00 6
## KnowledgeSources 0.00 1.000 1.000 0.800000 4.216370e-01 1.00 1.00 0
## KnowledgeDischarge 0.00 0.000 0.000 0.400000 5.163978e-01 1.00 1.00 0
## Scarcity_watershed NA NA NA NaN NA NA -Inf 10
## Scarcity_country NA 3.350 3.350 3.350000 NA 3.35 3.35 9
## Knowledge_Watershed 0.00 0.000 0.000 0.000000 0.000000e+00 0.00 0.00 0
## Knowledge_Facilities 0.00 0.000 0.000 0.200000 4.216370e-01 0.00 1.00 0
## Importance_num NA 3.000 3.000 3.333333 8.164966e-01 3.00 5.00 4
## DetrimentalImpacts_num 0.00 0.000 0.000 0.100000 3.162278e-01 0.00 1.00 0
## Responsibility_num 1.00 1.000 2.000 1.800000 7.888106e-01 2.00 3.00 0
## > str(res)
## num [1:16, 1:8] 2011 2 -1.14 41080 NA ...
## - attr(*, "dimnames")=List of 2
## ..$ : chr [1:16] "Year" "Occurences" "Balance" "Withdrawal" ...
## ..$ : chr [1:8] "Min" "Q1" "Med" "Mean" ...
虽然 stargazer 在很多情况下是一个不错的选择，但我还是推荐 xtable 因为它的灵活性。
print(xtable(res),type="html")
我已经搜索了很长时间，但找不到真正简单的方法。 我有一个仅由数值组成的 df，我想从我的 df 中创建一个汇总矩阵。
DF
V1 V2 V3 V4 V5 ...
x1 y1 z1 1 c1
x2 NA z2 0 c2
x3 y3 z3 1 NA
...
V4本来是一个TRUE/FALSE变量转化为数值变量，应该可以正常使用。 我想获得以下内容：
N Mean SD Min 1st Median 3rd Max
V1
V2
V3
V4
V5
...
根据 N、平均值、SD、最小值、第 1、中值、第 3、最大值的值。
我试过简单的
as.data.frame(summary(DF))
我试过 stargazer 但由于某种原因不起作用（我猜是因为我有二进制变量）
stargazer(DF, type= "html", nobs = TRUE, type="html", mean.sd = TRUE, median = TRUE, iqr = TRUE,
+ digits=2, align=T)
我读了一些关于 qwraps2_summary_table 的文章。但是它们似乎都给出了与我正在寻找的 "design" 不同的 table。
我知道我也可以 运行 像这样的循环：
for(i in (1:length(DF)){
sum$N<-(????)
sum$Mean<-mean(DF[i])
....}
但这不是最好的解决方案。 有小费吗？谢谢！
这是我数据集的一部分
structure(list(Year = c(2011, 2012, 2013, 2014, 2015, 2016, 2017,
2018, 2018, 2011), Occurences = c(9L, 9L, 9L, 9L, 9L, 9L, 9L,
9L, 2L, 9L), Balance = c(-1.14, 1.05, -1.06, 1.01, 1.01, 1.01,
-1.09, -1, -1.04, -1.03), Withdrawal = c(43200, 41080, 43400,
43183, 42600, 42100, 45900, 46000, 3892008, 48374), Verification_SA = c(NA,
NA, NA, NA, 1, 1, NA, 1, 1, NA), Classification_num = c(NA, NA,
NA, NA, 3, 2, NA, 4, 4, NA), Interaction_Verification_Classification = c(NA,
NA, NA, NA, 3, 2, NA, 4, 4, NA), KnowledgeSources = c(1, 1, 1,
0, 1, 1, 1, 1, 1, 0), KnowledgeDischarge = c(0, 0, 0, 0, 0, 1,
1, 1, 1, 0), Scarcity_watershed = c(NA_real_, NA_real_, NA_real_,
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_
), Scarcity_country = c(NA, NA, NA, NA, NA, NA, NA, NA, 3.35,
NA), Knowledge_Watershed = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Knowledge_Facilities = c(0,
0, 0, 0, 0, 0, 0, 0, 1, 1), Importance_num = c(NA, NA, NA, 3,
3, 3, 3, 3, 5, NA), DetrimentalImpacts_num = c(0, 0, 1, 0, 0,
0, 0, 0, 0, 0), Responsibility_num = c(1, 1, 1, 2, 2, 2, 2, 3,
3, 1)), row.names = c(NA, -10L), class = c("tbl_df", "tbl", "data.frame"
))
如果以后有人发现这个问题，请根据@camille 的建议检查 NA 的数量。
library(psych)
my_summary <- do.call(rbind,lapply(DF,psych::describe,quant=c(0.25,0.75)))
my_summary
# vars n mean sd median trimmed mad min max range skew kurtosis se Q0.25 Q0.75
#Year 1 10 2014.50 2.72 2014.50 2014.50 3.71 2011.00 2018.00 7.00 0.00 -1.74 0.86 2012.25 2016.75
#Occurences 1 10 8.30 2.21 9.00 9.00 0.00 2.00 9.00 7.00 -2.28 3.57 0.70 9.00 9.00
#Balance 1 10 -0.23 1.07 -1.02 -0.27 0.15 -1.14 1.05 2.19 0.35 -2.05 0.34 -1.06 1.01
#Withdrawal 1 10 428784.50 1216854.64 43300.00 44344.62 2535.25 41080.00 3892008.00 3850928.00 2.28 3.57 384803.22 42745.75 45975.00
#Verification_SA 1 4 1.00 0.00 1.00 1.00 0.00 1.00 1.00 0.00 NaN NaN 0.00 1.00 1.00
#Classification_num 1 4 3.25 0.96 3.50 3.25 0.74 2.00 4.00 2.00 -0.32 -2.08 0.48 2.75 4.00
...
这是我的原始解决方案，使用 data.table。
library(data.table)
my_summary <- rbindlist(lapply(DF, function(x){
as.data.frame(t(c(
summary(x),
SD = sd(x,na.rm=TRUE),
N = sum(!is.na(x)))))
})
, fill = TRUE, use.names = TRUE,idcol="Variable")
my_summary
# Variable Min. 1st Qu. Median Mean 3rd Qu. Max. SD N NA's
# 1: Year 2011.00 2012.250 2014.500 2014.500000 2016.75 2018.00 2.718251e+00 10 NA
# 2: Occurences 2.00 9.000 9.000 8.300000 9.00 9.00 2.213594e+00 10 NA
# 3: Balance -1.14 -1.055 -1.015 -0.228000 1.01 1.05 1.074800e+00 10 NA
# 4: Withdrawal 41080.00 42745.750 43300.000 428784.500000 45975.00 3892008.00 1.216855e+06 10 NA
# 5: Verification_SA 1.00 1.000 1.000 1.000000 1.00 1.00 0.000000e+00 4 6
# 6: Classification_num 2.00 2.750 3.500 3.250000 4.00 4.00 9.574271e-01 4 6
我们可以使用map循环遍历DF，获取summary统计数据，将其转换为data.frame，创建。 tibble 中的 'SD' 和 'N' 列以创建带有后缀的单个 data.frame 输出（_dfr in map）
library(purrr)
library(dplyr)
map_dfr(DF, ~ tibble(SD = sd(.x, na.rm = TRUE),
N = sum(!is.na(.x)), as.data.frame.list(base::summary(.x))),
.id = 'Variable')
# A tibble: 16 x 10
# SD N Min. X1st.Qu. Median Mean X3rd.Qu. Max. NA.s Variable
# * <dbl> <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr>
# 1 2.72e+0 10 2011 2012. 2014. 2.01e+3 2017. 2.02e3 NA Year
# 2 2.21e+0 10 2 9 9 8.30e+0 9 9.00e0 NA Occurences
# 3 1.07e+0 10 -1.14 -1.06 -1.02 -2.28e-1 1.01 1.05e0 NA Balance
# 4 1.22e+6 10 41080 42746. 43300 4.29e+5 45975 3.89e6 NA Withdrawal
# 5 0. 4 1 1 1 1.00e+0 1 1.00e0 6 Verification_SA
# 6 9.57e-1 4 2 2.75 3.5 3.25e+0 4 4.00e0 6 Classification_num
# 7 9.57e-1 4 2 2.75 3.5 3.25e+0 4 4.00e0 6 Interaction_Verificatio…
# 8 4.22e-1 10 0 1 1 8.00e-1 1 1.00e0 NA KnowledgeSources
# 9 5.16e-1 10 0 0 0 4.00e-1 1 1.00e0 NA KnowledgeDischarge
#10 NA 0 NA NA NA NaN NA NA 10 Scarcity_watershed
#11 NA 1 3.35 3.35 3.35 3.35e+0 3.35 3.35e0 9 Scarcity_country
#12 0. 10 0 0 0 0. 0 0. NA Knowledge_Watershed
#13 4.22e-1 10 0 0 0 2.00e-1 0 1.00e0 NA Knowledge_Facilities
#14 8.16e-1 6 3 3 3 3.33e+0 3 5.00e0 4 Importance_num
#15 3.16e-1 10 0 0 0 1.00e-1 0 1.00e0 NA DetrimentalImpacts_num
#16 7.89e-1 10 1 1 2 1.80e+0 2 3.00e0 NA Responsibility_num
要以 Ian Campbell 的回答为基础，在需要时不要害怕构建汇总函数。
summaryfn <- function(x){
c(min(x),
quantile(x,0.25,na.rm=TRUE),
quantile(x,0.5,na.rm=TRUE),
mean(x,na.rm=TRUE),
sd(x, na.rm=TRUE),
quantile(x,0.75,na.rm=TRUE),
max(x,na.rm=TRUE),
sum(is.na(x)))
}
res <- do.call(rbind,lapply(df,summaryfn))
colnames(res) <- c("Min","Q1","Med","Mean","Sd","Q3","Max","NAs")
## > res
## Min Q1 Med Mean Sd Q3 Max NAs
## Year 2011.00 2012.250 2014.500 2014.500000 2.718251e+00 2016.75 2018.00 0
## Occurences 2.00 9.000 9.000 8.300000 2.213594e+00 9.00 9.00 0
## Balance -1.14 -1.055 -1.015 -0.228000 1.074800e+00 1.01 1.05 0
## Withdrawal 41080.00 42745.750 43300.000 428784.500000 1.216855e+06 45975.00 3892008.00 0
## Verification_SA NA 1.000 1.000 1.000000 0.000000e+00 1.00 1.00 6
## Classification_num NA 2.750 3.500 3.250000 9.574271e-01 4.00 4.00 6
## Interaction_Verification_Classification NA 2.750 3.500 3.250000 9.574271e-01 4.00 4.00 6
## KnowledgeSources 0.00 1.000 1.000 0.800000 4.216370e-01 1.00 1.00 0
## KnowledgeDischarge 0.00 0.000 0.000 0.400000 5.163978e-01 1.00 1.00 0
## Scarcity_watershed NA NA NA NaN NA NA -Inf 10
## Scarcity_country NA 3.350 3.350 3.350000 NA 3.35 3.35 9
## Knowledge_Watershed 0.00 0.000 0.000 0.000000 0.000000e+00 0.00 0.00 0
## Knowledge_Facilities 0.00 0.000 0.000 0.200000 4.216370e-01 0.00 1.00 0
## Importance_num NA 3.000 3.000 3.333333 8.164966e-01 3.00 5.00 4
## DetrimentalImpacts_num 0.00 0.000 0.000 0.100000 3.162278e-01 0.00 1.00 0
## Responsibility_num 1.00 1.000 2.000 1.800000 7.888106e-01 2.00 3.00 0
## > str(res)
## num [1:16, 1:8] 2011 2 -1.14 41080 NA ...
## - attr(*, "dimnames")=List of 2
## ..$ : chr [1:16] "Year" "Occurences" "Balance" "Withdrawal" ...
## ..$ : chr [1:8] "Min" "Q1" "Med" "Mean" ...
虽然 stargazer 在很多情况下是一个不错的选择，但我还是推荐 xtable 因为它的灵活性。
print(xtable(res),type="html")