计算2列之一中存在空白的实例数＆＃x2b; r

发布于 2025-02-01 00:28:39 字数 272 浏览 3 评论 0原文

我需要测试一个数据框以完成记录 - 要完成记录，必须在两个列中进行条目。在示例df中，您将看到9个条目中的2个中包含一个行中的空白。

df <- data.frame(a = c(1,2,"",4,5,6,7,"",8,9),
                 b = c(9,5,2,7,5,"",3,"",6,8))

所需的输出将是2或7识别完整或相反不完整的记录数量的计数。两者都是空白的实例。

原文

I need to test a dataframe for completion of records - for the records to be complete an entry must be made in both columns. In in the example df below you will see that 2 of the 9 entries contain blanks in one of the rows.

df <- data.frame(a = c(1,2,"",4,5,6,7,"",8,9),
                 b = c(9,5,2,7,5,"",3,"",6,8))

The desired output would be a count of 2 or 7 identifying the number of records that are complete or conversely incomplete. Instances where both are blanks would not be considered in the count.

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傲世九天 2025-02-08 00:28:39

它可以是逻辑矩阵的sum - 将'df'转换为逻辑矩阵（df！=''），获取> rowsums并检查sum（true - ＆gt; 1和false - ＆gt; 0）等于列的列数，以返回n1 n1 < /code>并从排行的总数中减去n2

n1 <- sum(rowSums(df != '') == ncol(df))
n1
[1] 7
n2 <- nrow(df) - n1
# or if there are some other cases with `NA` etc
n2 <- sum(rowSums(df =='') == 1, na.rm = TRUE)

It could be sum of logical matrix - convert the 'df' to logical matrix (df != ''), get the rowSums and check whether the sum (TRUE -> 1 and FALSE -> 0) is equal to number of columns to return n1 and subtract from the total number of rows to get n2

n1 <- sum(rowSums(df != '') == ncol(df))
n1
[1] 7
n2 <- nrow(df) - n1
# or if there are some other cases with `NA` etc
n2 <- sum(rowSums(df =='') == 1, na.rm = TRUE)

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