英文:
Transform a dataframe with column value as column names in R
问题
我理解你的需求，你想将原始的数据框按照特定的方式进行转换，使其具有类似于示例中的结构。你提到尝试使用pivot_wider，但似乎没有成功。以下是你提供的代码的修改版本：
# 使用pivot_wider重新尝试
df_transformed <- df %>%
pivot_wider(names_from = date,
values_from = c(price, num_floors),
names_glue = "{.value}_{date}",
values_fill = list(price = NaN, num_floors = NaN))
这将按照你描述的方式转换数据框，每列包含给定日期的价格和楼层数。每列中，前两行是第一个房屋的数据，接下来两行是第二个房屋的数据，并使用NaN填充缺失的条目。
英文:
I have a dataframe like this
date price num_floors house
1 2023-01-01 94.30076 3 A
2 2023-01-01 95.58771 2 B
3 2023-01-02 102.78559 1 C
4 2023-01-03 93.29053 3 D
and I want to change it, so that each column contains the prices and num_floors for all houses for a given date. For one column, the first two rows of a column refer to the first house, the next two to the second house. The remaining entries without data are filled with the missing value NaN.
Now, I want to transform the above dataframe, so that it has a similar structure as the following:
2023-01-01 2023-01-02 2023-01-03
1 94.30076 102.78559 93.29053
2 3 1 3
3 95.58771 NA NA
4 2 NA NA
Each column contains the prices and num_floors for all houses for a given date. For one column, the first two rows of a column refer to the first house, the next two to the second house. The remaining entries without data are filled with the missing value NaN.
I tried with pivot_wider. It does not work:
# Pivot the dataframe
df_transformed <- df %>%
pivot_wider(names_from = date,
values_from = c(price, num_floors),
values_fill = NaN)
答案1
得分: 1
以下是翻译好的内容：
"首先，您可以将数据转换为长格式，然后在添加行号后再转换为宽格式。"
"输出："
| `2023-01-01` | `2023-01-02` | `2023-01-03` |
| ------------ | ------------ | ------------ |
| <dbl> | <dbl> | <dbl> |
| 94.3 | 103. | 93.3 |
| 3 | 1 | 3 |
| 95.6 | NA | NA |
| 2 | NA | NA |
（我要注意的是，这种转换方式似乎不太可能是表示您的数据的最佳方式）。
英文:
You can pivot to long format first, and then go wide after adding a row number.
pivot_wider(
pivot_longer(df, -date) %>% mutate(n=row_number(), .by=date),
id_cols = n, names_from=date, values_from = value
) %>% select(-n)
Output:
`2023-01-01` `2023-01-02` `2023-01-03`
<dbl> <dbl> <dbl>
1 94.3 103. 93.3
2 3 1 3
3 95.6 NA NA
4 2 NA NA
(I will note that this transformation seem unlikely to be the optimal way to represent your data).
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