排序数据框后更新索引

Take the following data-frame:

x = np.tile(np.arange(3),3)
y = np.repeat(np.arange(3),3)
df = pd.DataFrame({"x": x, "y": y})

   x  y
0  0  0
1  1  0
2  2  0
3  0  1
4  1  1
5  2  1
6  0  2
7  1  2
8  2  2

I need to sort it by x first, and only second by y:

df2 = df.sort(["x", "y"])

   x  y
0  0  0
3  0  1
6  0  2
1  1  0
4  1  1
7  1  2
2  2  0
5  2  1
8  2  2

How can I change the index such that it is ascending again. I.e. how do I get this:

   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2

I have tried the following. Unfortunately, it doesn’t change the index at all:

df2.reindex(np.arange(len(df2.index)))

You can reset the index using to get back a default index of 0, 1, 2, …, n-1 (and use drop=True to indicate you want to drop the existing index instead of adding it as an additional column to your dataframe):

In [19]: df2 = df2.reset_index(drop=True)

In [20]: df2
Out[20]:
   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2

df.sort() is deprecated, use df.sort_values(...): https://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.sort_values.html

Then follow joris’ answer by doing df.reset_index(drop=True)

Since pandas 1.0.0 has a new parameter ignore_index which does exactly what you need:

In [1]: df2 = df.sort_values(by=['x','y'],ignore_index=True)

In [2]: df2
Out[2]:
   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2

You can set new indices by using set_index:

df2.set_index(np.arange(len(df2.index)))

Output:

   x  y
0  0  0
1  0  1
2  0  2
3  1  0
4  1  1
5  1  2
6  2  0
7  2  1
8  2  2