pandas 更改 DataFrame 最后一行中的元素
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Change an element in the last row of a DataFrame
提问by Mike
I set up a simple DataFrame in pandas:
我在 Pandas 中设置了一个简单的 DataFrame：
a = pandas.DataFrame([[1,2,3], [4,5,6], [7,8,9]], columns=['a','b','c'])
>>> print a
a b c
0 1 2 3
1 4 5 6
2 7 8 9
I would like to be able to alter a single element in the last row of. In pandas==0.13.1 I could use the following:
我希望能够更改最后一行中的单个元素。在 pandas==0.13.1 我可以使用以下内容：
a.iloc[-1]['a'] = 77
>>> print a
a b c
0 1 2 3
1 4 5 6
2 77 8 9
but after updating to pandas==0.14.1, I get the following warning when doing this:
但是在更新到 pandas==0.14.1 后，我在执行此操作时收到以下警告：
SettingWithCopyWarning: A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_index,col_indexer] = value instead
The problem of course being that -1 is not an index of a, so I can't use loc. As the warning indicates, I have not changed column 'a'of the last row, I've only altered a discarded local copy.
问题当然是 -1 不是 的索引a，所以我不能使用loc. 正如警告所示，我没有更改'a'最后一行的列，我只更改了丢弃的本地副本。
How do I do this in the newer version of pandas? I realize I could use the index of the last row like:
我如何在较新版本的Pandas中做到这一点？我意识到我可以使用最后一行的索引，例如：
a.loc[2,'a'] = 77
But I'll be working with tables where multiple rows have the same index, and I don't want to reindex my table every time. Is there a way to do this without knowing the index of the last row before hand?
但是我将使用多行具有相同索引的表，并且我不想每次都重新索引我的表。有没有办法在不事先知道最后一行的索引的情况下做到这一点？
采纳答案by Mike
Alright I've found a way to solve this problem without chaining, and without worrying about multiple indices.
好吧，我找到了一种无需链接即可解决此问题的方法，无需担心多个索引。
a.iloc[-1, a.columns.get_loc('a')] = 77
>>> a
a b c
0 1 2 3
1 4 5 6
2 77 8 9
I wasn't able to use ilocbefore because I couldn't supply the column index as an int, but get_locsolves that problem. Thanks for the helpful comments everyone!
我iloc之前无法使用，因为我无法将列索引作为 int 提供，但get_loc解决了该问题。感谢大家的有益评论！
回答by brunostuyts
Taking elements from the solutions of @PallavBakshi and @Mike, the following works in Pandas >= 0.19
从@PallavBakshi 和@Mike 的解决方案中获取元素，以下在 Pandas >= 0.19 中有效
Just using iloc[-1, 'a] won't work as -1 is not in the index.
仅使用 iloc[-1, 'a] 将不起作用，因为 -1 不在索引中。
a.loc[a.index[-1], 'a']= 4.0
回答by PallavBakshi
For pandas 0.22,
对于Pandas 0.22，
a.at[a.index[-1], 'a'] = 77
this is just one of the ways.
这只是其中一种方式。
