Question

首先，让我们从DataFrame 1（DF1）开始：

DF1 <- data.frame(c("06/19/2016", "06/20/2016", "06/21/2016", "06/22/2016", 
                    "06/23/2016", "06/19/2016", "06/20/2016", "06/21/2016",
                    "06/22/2016", "06/23/2016"),
                  c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
                  c(149, 150, 151, 152, 155, 84, 83, 80, 81, 97),
                  c(101, 102, 104, 107, 99, 55, 55, 56, 57, 58),
                  c("MTL", "MTL", "MTL", "MTL", "MTL", "NY", "NY", 
                    "NY", "NY", "NY"))
colnames(DF1) <- c("date", "id", "sales", "cost", "city")

我也有DataFrame 2（DF2）：

DF2 <- data.frame(c("06/19/2016", "06/27/2016", "06/22/2016", "06/23/2016"),
                  c(1, 1, 2, 2),
                  c(9999, 8888, 777, 555),
                  c("LON", "LON", "QC", "QC"))
colnames(DF2) <- c("date", "id", "sales", "city")

对于DF1中的每一行，我必须查看DF2中是否有一行具有相同的日期和ID。如果是，我必须用DF2中的值替换DF1中的值。

DF2的列数始终少于DF1。如果列不在DF2中，我必须保留该特定列的DF1中的原始值。

最终输出是这样的：

results <- data.frame(c("06/19/2016", "06/20/2016", "06/21/2016", "06/22/2016",
                        "06/23/2016", "06/19/2016", "06/20/2016", "06/21/2016",
                        "06/22/2016", "06/23/2016"),
                      c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2),
                      c(9999, 150, 151, 152, 155, 84, 83, 80, 777, 555),
                      c(101, 102, 104, 107, 99, 55, 55, 56, 57, 58),
                      c("LON", "MTL", "MTL", "MTL", "MTL", "NY", "NY", 
                        "NY", "QC", "QC"))
colnames(results) <- c("date", "id", "sales", "cost", "city")

你有什么建议吗？

Answer 1

您可以使用data.table - 包的连接功能：

library(data.table)
setDT(DF1)
setDT(DF2)

DF1[DF2, on = .(date, id), `:=` (city = i.city, sales = i.sales)]

给出：

> DF1
          date id sales cost city
 1: 06/19/2016  1  9999  101  LON
 2: 06/20/2016  1   150  102  MTL
 3: 06/21/2016  1   151  104  MTL
 4: 06/22/2016  1   152  107  MTL
 5: 06/23/2016  1   155   99  MTL
 6: 06/19/2016  2    84   55   NY
 7: 06/20/2016  2    83   55   NY
 8: 06/21/2016  2    80   56   NY
 9: 06/22/2016  2   777   57   QC
10: 06/23/2016  2   555   58   QC

如果两个数据集中有很多列，则更容易使用mget而不是键入所有列名称。对于问题中使用的数据，它看起来像：

DF1[DF2, on = .(date, id), names(DF2)[3:4] := mget(paste0("i.", names(DF2)[3:4]))]

Answer 2

df <- merge(DF1, DF2, by = c("date", "id"), all.x=TRUE)

tmp1 <- df[is.na(df$sales.y) & is.na(df$city.y),]
tmp1$sales.y <- NULL
tmp1$city.y <- NULL
names(tmp1)[names(tmp1) == "sales.x"] <- "sales"
names(tmp1)[names(tmp1) == "city.x"] <- "city"

tmp2 <- df[!is.na(df$sales.y) & !is.na(df$city.y),]
tmp2$sales.x <- NULL
tmp2$city.x <- NULL
names(tmp2)[names(tmp2) == "sales.y"] <- "sales"
names(tmp2)[names(tmp2) == "city.y"] <- "city"

results <- rbindlist(list(tmp1,tmp2), use.names= TRUE, fill = TRUE)

See the result

Answer 3

column

将SELECT DISTINCT (((p.raw_data ->> 'company') :: JSONB) ->> 'userId') :: INTEGER FROM schema.projects p WHERE p.raw_data :: JSONB ? 'company' AND (p.raw_data ->> 'company') :: JSONB ? 'userId' AND (((p.raw_data ->> 'company') :: JSONB) ->> 'userId') :: INTEGER > 0 AND p.is_deleted = FALSE LIMIT 20; -- Returns 20 results替换为您的变量。

根据其他数据框替换特定值

3 个答案: