找到多个日期列中的最小和最大日期,使用R。

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英文:

Find Min and Max date from multiple date columns using R

问题

# 创建新的列 maxdate 和 mindate
df$maxdate <- apply(df[1:4], 1, max, na.rm = TRUE)
df$mindate <- apply(df[1:4], 1, min, na.rm = TRUE)
英文:

I would like to create a new column called maxdate and mindate from a list of date columns assuming 4 date columns and has missing values.

The solution below only gives the max/min of the rows of the columns. I am interested in finding max/min date across the columns.

df$maxdate &lt;- apply (df[1:4], 1, max, na.rm = TRUE)
df &lt;- data.frame(
  col1 = c(&quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;),
  col2 = c(&quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;),
  col3 = c(&quot;12/09/1961&quot;, &quot;10/03/1995&quot;, &quot;30/03/1992&quot;, &quot;25/05/1992&quot;, &quot;25/05/1992&quot;),
  col4 = c(&quot;01/01/1930&quot;, &quot;01/01/1939&quot;, &quot;01/01/1942&quot;, &quot;01/01/1936&quot;, &quot;01/01/1937&quot;)
)

sample data

col1          col2        col3      col4

11/09/1999	01/01/2000	12/09/1961	01/01/1930
11/09/1999	01/01/2000	10/03/1995	01/01/1939
11/09/1999	01/01/2000	30/03/1992	01/01/1942
11/09/1999	01/01/2000	25/05/1992	01/01/1936
11/09/1999	01/01/2000	25/05/1992	01/01/1937

答案1

得分: 0

库(dplyr)

df = 数据框(date1 = c("2023-05-11", "2023-04-12","2023-07-13","2023-01-14","2023-05-15"),
            date2 = c("2023-04-11", "2023-07-12","2023-09-13","2023-05-14","2023-12-15"),
            date3 = c("2023-08-11", "2023-06-12","2023-08-13","2023-08-14","2023-05-15"),
            date4 = c("2023-01-11", "2023-05-12","2023-05-13","2023-12-14","2023-05-15"))

df <- df  %>% mutate_all(as.Date)

# 编辑:删除了rowwise并添加了na.rm=TRUE,因为您似乎希望从所有行中获取最大值,忽略NAs?
df <- df %>%  mutate(max_date = max(date1, date2,date3,date4, na.rm=TRUE))

df
英文:
library(dplyr)

df = data.frame(date1 = c(&quot;2023-05-11&quot;, &quot;2023-04-12&quot;,&quot;2023-07-13&quot;,&quot;2023-01-14&quot;,&quot;2023-05-15&quot;),
                date2 = c(&quot;2023-04-11&quot;, &quot;2023-07-12&quot;,&quot;2023-09-13&quot;,&quot;2023-05-14&quot;,&quot;2023-12-15&quot;),
                date3 = c(&quot;2023-08-11&quot;, &quot;2023-06-12&quot;,&quot;2023-08-13&quot;,&quot;2023-08-14&quot;,&quot;2023-05-15&quot;),
                date4 = c(&quot;2023-01-11&quot;, &quot;2023-05-12&quot;,&quot;2023-05-13&quot;,&quot;2023-12-14&quot;,&quot;2023-05-15&quot;))

df &lt;- df  %&gt;% mutate_all(as.Date)

# edit: removed rowwise and added na.rm=TRUE, as you seem to want the max from all rows, disregarding NAs?
df &lt;- df %&gt;%  mutate(max_date = max(date1, date2,date3,date4, na.rm=TRUE))

df

答案2

得分: 0

翻译的部分如下:

df <- data.frame(
  col1 = c("11/09/1999", "11/09/1999", "11/09/1999", "11/09/1999", "11/09/1999"),
  col2 = c("01/01/2000", "01/01/2000", "01/01/2000", "01/01/2000", "01/01/2000"),
  col3 = c("12/09/1961", "10/03/1995", "30/03/1992", "25/05/1992", "25/05/1992"),
  col4 = c("01/01/1930", "01/01/1939", "01/01/1942", "01/01/1936", "01/01/1937")
)

maxdate <- lapply(df, function (x) max(x, na.rm = TRUE))

> maxdate
$col1
[1] "11/09/1999"

$col2
[1] "01/01/2000"

$col3
[1] "30/03/1992"

$col4
[1] "01/01/1942"

请注意,由于行数与列数不相等,因此我未将其分配给列,因此您不能将其添加到数据框中。

英文:

Like this?

df &lt;- data.frame(
  col1 = c(&quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;, &quot;11/09/1999&quot;),
  col2 = c(&quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;, &quot;01/01/2000&quot;),
  col3 = c(&quot;12/09/1961&quot;, &quot;10/03/1995&quot;, &quot;30/03/1992&quot;, &quot;25/05/1992&quot;, &quot;25/05/1992&quot;),
  col4 = c(&quot;01/01/1930&quot;, &quot;01/01/1939&quot;, &quot;01/01/1942&quot;, &quot;01/01/1936&quot;, &quot;01/01/1937&quot;)
)


maxdate &lt;- lapply(df, function (x) max(x, na.rm = TRUE))




&gt; maxdate
$col1
[1] &quot;11/09/1999&quot;

$col2
[1] &quot;01/01/2000&quot;

$col3
[1] &quot;30/03/1992&quot;

$col4
[1] &quot;01/01/1942&quot;

Note that I did not assign it to a column as the number of rows != the number of columns so you cannot add it to your dataframe.

huangapple
  • 本文由 发表于 2023年5月17日 17:49:47
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