python - np.where() 如果条件失败则什么都不做

Question

我有一个来 self 的数据框的样本:

       Created      Insert Time   MatchKey              In Previous    New Type
18593  2016-08-12   2018-02-19    LXGS090393APIN040640       No        New Existing
5517   2016-08-12   2018-02-19    LIN380076CI166203726       No        New Existing
2470   2018-02-12   2018-02-19    CI164414649APIN160672      No        New Existing
13667  2016-08-12   2018-02-19    LIN257400APIN015446       Yes        New Existing
10998  2016-08-12   2018-02-19    LXSV225786APIN158860      Yes        New Existing
20149  2016-08-12   2018-02-19    LIN350167APIN158284       Yes        New Existing
20143  2016-08-12   2018-02-19    LIN350167APIN161348       Yes        New Existing
30252  2016-08-12   2018-02-19    LXGS120737APIN153339      Yes        New Existing
12583  2016-08-09   2018-02-19    WIN556410APIN157186       Yes        New Existing
28591  2018-05-03   2018-02-19    CI195705185APIN009076      No        New Created

我想以一种方式替换 New Type 列中的值，如果条件失败，函数什么都不做:

current['New Type'] = np.where(current['In Previous']=='Yes','In Previous',pass)

但显然它会导致语法错误，因为 np.where() 不处理 pass:

File "<ipython-input-9-7f68cda12cbe>", line 1
current['New Type'] = np.where(current['In Previous']=='Yes','In Previous',pass)

                                                                          ^
SyntaxError: invalid syntax

有什么替代方案可以实现同样的目标？

Answer 1

只返回列而不是 pass 这等同于当条件为 False 时什么都不做:

current['New Type'] = np.where(current['In Previous']=='Yes','In Previous',current['New Type'] )

或者您可以只屏蔽这些行:

current['New Type'] = current.loc[current['In Previous']=='Yes', 'In Previous']