如何使用條件打印某個數據幀的某些行

Question

從下面的data.frame中，我想使用print（）函數來簡單地打印出具有「red」的行的行信息（日期和溫度） 「警告變量。我嘗試過索引，但沒有運氣！ Tia

   date temp warnings 
      <dttm> <dbl> <chr> 
1 2017-04-16 10:00:00 26.3 black 
2 2017-04-17 10:00:00 25.7 black 
3 2017-04-18 10:00:00 25.4 black 
4 2017-04-19 10:00:00 25.6 black 
5 2017-04-20 10:00:00 25.9 black 
6 2017-04-21 10:00:00 26.1 black 
7 2017-04-22 10:00:00 27.8 red 
8 2017-04-23 10:00:00 26.3 black 
9 2017-04-24 10:00:00 26.1 black 
10 2017-04-25 10:00:00 26.6 black 

謝謝大家！我已經有一個玩轉你的建議，並設法打印雖然有反正這個了，而不是打印列標題？

df[df$warnings == 'red', c('date', 'temp')] 
# A tibble: 2 x 2 
       date temp 
       <dttm> <dbl> 
1 2017-05-27 10:00:00 27.8 
2 2017-05-29 10:00:00 27.7 

Answer 1

> library("readr") 
> red2=read_csv("red.csv") 
Parsed with column specification: 
cols(
    date = col_character(), 
    temp = col_double(), 
    warning = col_character() 
) 
> red2 
# A tibble: 10 x 3 
       date temp warning 
      <chr> <dbl> <chr> 
1 4/16/2017 10:00 26.3 black 
2 4/17/2017 10:00 25.7 black 
3 4/18/2017 10:00 25.4 black 
4 4/19/2017 10:00 25.6 black 
5 4/20/2017 10:00 25.9 black 
6 4/21/2017 10:00 26.1 black 
7 4/22/2017 10:00 27.8  red 
8 4/23/2017 10:00 26.3 black 
9 4/24/2017 10:00 26.1 black 
10 4/25/2017 10:00 26.6 black 
> library(lubridate) 

Attaching package: ‘lubridate’ 

The following object is masked from ‘package:base’: 

    date 

> red2$date=mdy_hm(red2$date) 
> red2 
# A tibble: 10 x 3 
        date temp warning 
       <dttm> <dbl> <chr> 
1 2017-04-16 10:00:00 26.3 black 
2 2017-04-17 10:00:00 25.7 black 
3 2017-04-18 10:00:00 25.4 black 
4 2017-04-19 10:00:00 25.6 black 
5 2017-04-20 10:00:00 25.9 black 
6 2017-04-21 10:00:00 26.1 black 
7 2017-04-22 10:00:00 27.8  red 
8 2017-04-23 10:00:00 26.3 black 
9 2017-04-24 10:00:00 26.1 black 
10 2017-04-25 10:00:00 26.6 black 

只選擇那些警告紅色行，由他們列數

>  red2[red2$warning=="red",1:2] 
# A tibble: 1 x 2 
       date temp 
       <dttm> <dbl> 
1 2017-04-22 10:00:00 27.8 

我們解決大數據問題只選擇兩列，我模擬的數據要大些

> red3=red2[rep(seq_len(nrow(red2)), each=2000000),] 
> str(red3) 
Classes ‘tbl_df’, ‘tbl’ and 'data.frame':  20000000 obs. of 3 variables: 
$ date : POSIXct, format: "2017-04-16 10:00:00" "2017-04-16 10:00:00" ... 
$ temp : num 26.3 26.3 26.3 26.3 26.3 26.3 26.3 26.3 26.3 26.3 ... 
$ warning: chr "black" "black" "black" "black" ... 
> system.time(red3[red3$warning=="red",1:2]) 
    user system elapsed 
    0.17 0.06 0.24 

現在打印出沒有列標題的數據，我們使用矩陣與dimnames = NULL從How do you delete the header in a dataframe?

> out=red2[red2$warning=="red",1:2]> out.print <- function(dat) print(matrix(as.matrix(out),ncol=ncol(out),dimnames=NULL),quote=F) 
> out.print(out) 
    [,1]    [,2] 
[1,] 2017-04-22 10:00:00 27.8 

Answer 2

我用這樣的任務包'dplyr'。

install.packages(dplyr) 
library(dplyr) 
filter(my_df, warnings =='red') 

其中my_df是您的表的名稱。