使用pandas重新排列列：是否有与dplyr的select（...，everything（））等效的选项？

Question

我想要重新排列DataFrame中的列，将一些列放在前面，然后将所有其他列放在后面。

使用R的dplyr，代码如下：

library(dplyr)

df = tibble(col1 = c("a", "b", "c"),
            id = c(1, 2, 3),
            col2 = c(2, 4, 6),
            date = c("1 Feb", "2 Feb", "3 Feb"))

df2 = select(df,
             id, date, everything())

很简单。使用Python的pandas，我尝试了以下内容：

import pandas as pd

df = pd.DataFrame({
    "col1": ["a", "b", "c"],
    "id": [1, 2, 3],
    "col2": [2, 4, 6],
    "date": ["1 Feb", "2 Feb", "3 Feb"]
    })

# using sets
cols = df.columns.tolist()
cols_1st = {"id", "date"}
cols = set(cols) - cols_1st
cols = list(cols_1st) + list(cols)

# wrong column order
df2 = df[cols]

# using lists
cols = df.columns.tolist()
cols_1st = ["id", "date"]
cols = [c for c in cols if c not in cols_1st]
cols = cols_1st + cols

# right column order, but is there a better way?
df3 = df[cols]

使用 pandas 的方式更加繁琐，但我对此还比较陌生。有没有更好的方法呢？

- ardaar

Answer 1

您可以使用 df.drop：

>>> df = pd.DataFrame({
    "col1": ["a", "b", "c"],
    "id": [1, 2, 3],
    "col2": [2, 4, 6],
    "date": ["1 Feb", "2 Feb", "3 Feb"]
    })

>>> df

  col1  id  col2   date
0    a   1     2  1 Feb
1    b   2     4  2 Feb
2    c   3     6  3 Feb

>>> cols_1st = ["id", "date"]

>>> df[cols_1st + list(df.drop(cols_1st, 1))]

   id   date col1  col2
0   1  1 Feb    a     2
1   2  2 Feb    b     4
2   3  3 Feb    c     6

- Sayandip Dutta

---

谢谢你。我已经将你最初的 cols_1st = ["id", "date"] 更改为 cols_1st =[value for value in ["id", "date", "non existing column without exception"] if value in df.columns.tolist()]。适用于有很多列的数据框。 - phili_b

Answer 2

就像您在 R 中使用 datar 一样简单：

>>> from datar.all import c, f, tibble, select, everything
>>> df = tibble(col1 = c("a", "b", "c"),
...             id = c(1, 2, 3),
...             col2 = c(2, 4, 6),
...             date = c("1 Feb", "2 Feb", "3 Feb"))
>>>             
>>> df2 = select(df,
...              f.id, f.date, everything())
>>>              
>>> df2
       id     date     col1    col2
  <int64> <object> <object> <int64>
0       1    1 Feb        a       2
1       2    2 Feb        b       4
2       3    3 Feb        c       6

我是这个软件包的作者。如果您有任何问题，请随时提交问题。

- Panwen Wang

Answer 3

通常来说，R和Python Pandas之间的最佳转换是使用基础R，其遵循与逻辑索引相同的语义，例如在矢量上的逻辑索引，这里是列名。请注意以下与否定和in函数的相似之处：

# R 
mycols <- c("id", "date")
df2 <- df[c(mycols, colnames(df)[!colnames(df) %in% c(mycols)])]


# PANDAS (OLDER, NON-RECOMMENDED WAY)
mycols = ["id", "date"]
df2 = df[mycols + df.columns[~df.columns.isin(mycols)].tolist()]

# PANDAS (CURRENT, RECOMMENDED WAY WITH reindex)
df2 = df.reindex(mycols + df.columns[~df.columns.isin(mycols)].tolist(),
                 axis='columns')

- Parfait