如何在Pandas数据框中拆分一列元组? | 如何在Pandas数据框中拆分一列元组

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问题描述

我有一个Pandas数据框(这只是一小部分)

I have a Pandas dataframe (this is only a little piece)

>>> d1
   y norm test  y norm train  len(y_train)  len(y_test)  \
0    64.904368    116.151232          1645          549
1    70.852681    112.639876          1645          549

                                    SVR RBF  \
0   (35.652207342877873, 22.95533537448393)
1  (39.563683797747622, 27.382483096332511)

                                        LCV  \
0  (19.365430594452338, 13.880062435173587)
1  (19.099614489458364, 14.018867136617146)

                                   RIDGE CV  \
0  (4.2907610988480362, 12.416745648065584)
1    (4.18864306788194, 12.980833914392477)

                                         RF  \
0   (9.9484841581029428, 16.46902345373697)
1  (10.139848213735391, 16.282141345406522)

                                           GB  \
0  (0.012816232716538605, 15.950164822266007)
1  (0.012814519804493328, 15.305745202851712)

                                             ET DATA
0  (0.00034337162272515505, 16.284800366214057)  j2m
1  (0.00024811554516431878, 15.556506191784194)  j2m
>>>

我想拆分所有包含元组的列.例如，我想将列 LCV 替换为列 LCV-a 和 LCV-b .

I want to split all the columns that contain tuples. For example, I want to replace the column LCV with the columns LCV-a and LCV-b.

我该怎么做?

推荐答案

您可以通过在该列上执行 pd.DataFrame(col.tolist())来做到这一点:

You can do this by doing pd.DataFrame(col.tolist()) on that column:

In [2]: df = pd.DataFrame({'a':[1,2], 'b':[(1,2), (3,4)]})

In [3]: df
Out[3]:
   a       b
0  1  (1, 2)
1  2  (3, 4)

In [4]: df['b'].tolist()
Out[4]: [(1, 2), (3, 4)]

In [5]: pd.DataFrame(df['b'].tolist(), index=df.index)
Out[5]:
   0  1
0  1  2
1  3  4

In [6]: df[['b1', 'b2']] = pd.DataFrame(df['b'].tolist(), index=df.index)

In [7]: df
Out[7]:
   a       b  b1  b2
0  1  (1, 2)   1   2
1  2  (3, 4)   3   4

注意:在早期版本中，此答案建议使用 df ['b'].apply(pd.Series)而不是 pd.DataFrame(df ['b'].tolist()，index = df.index).这样做也很好(因为它使每个元组组成一个Series，然后将其视为一个数据帧的一行)，但是它比 tolist 版本更慢/使用更多的内存，如其他答案(感谢到denfromufa ).

Note: in an earlier version, this answer recommended to use df['b'].apply(pd.Series) instead of pd.DataFrame(df['b'].tolist(), index=df.index). That works as well (because it makes a Series of each tuple, which is then seen as a row of a dataframe), but it is slower / uses more memory than the tolist version, as noted by the other answers here (thanks to denfromufa).

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