python - 用 Pandas : Is there an equivalent to dplyr's select(. ..，一切())重新排列列？

Question

我正在尝试重新排列 DataFrame 中的列，首先放置几列，然后放置所有其他列。

与R的dplyr ，这看起来像:

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())

简单。使用 Python 的 pandas ，这是我尝试过的:

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]

pandas方式更乏味，但我对此相当陌生。有没有更好的办法？

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