pandas 熊猫：求和两行数据帧而不重新排列数据帧？
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pandas: sum two rows of dataframe without rearranging dataframe?
提问by ale19
I have a dataframe and I'm trying to sum two rows without messing up the order of the rows.
我有一个数据框，我试图在不弄乱行顺序的情况下对两行求和。
> test = {'counts' : pd.Series([10541,4143,736,18,45690], index=['Daylight','Dawn','Other / unknown','Uncoded & errors','Total']), 'percents' : pd.Series([23.07,9.07,1.61,0.04,100], index=['Daylight','Dawn','Other / unknown','Uncoded & errors','Total'])}
> testdf = pd.DataFrame(test)
counts percents
Daylight 10541 23.07
Dawn 4143 9.07
Other / unknown 736 1.61
Uncoded & errors 18 0.04
Total 45690 100.00
I want this output:
我想要这个输出：
counts percents
Daylight 10541 23.07
Dawn 4143 9.07
Other / unknown 754 1.65 <-- sum of 'other/unknown' and 'uncoded & errors'
Total 45690 100.00
This is as close as I've been able to get:
这是我所能得到的最接近的：
> sum_ = testdf.loc[['Other / unknown', 'Uncoded & errors']].sum().to_frame().transpose()
counts percents
0 754.00 1.65
> sum_ = sum_.rename(index={0: 'Other / unknown'})
counts percents
Other / unknown 754.00 1.65
> testdf.drop(['Other / unknown', 'Uncoded & errors'],inplace=True)
> testdf = testdf.append(sum_)
Daylight 10541 23.07
Dawn 4143 9.07
Total 45690 100
Other / unknown 754 1.65
But this does not preserve the order of the original rows
但这不会保留原始行的顺序
I could insert the row by slicing the dataframe and inserting the sum_ row between 'Dawn' and 'Total', but that will not work if the row labels ever change, or if the order of the rows change, etc. (this is an annual brochure so the table design might change from year to year), so I'm trying to do this robustly.
我可以通过切片数据框并在 'Dawn' 和 'Total' 之间插入 sum_ 行来插入行，但是如果行标签发生变化，或者行的顺序发生变化等，这将不起作用（这是一个年度小册子，因此表格设计可能会逐年变化），所以我正在努力做到这一点。
回答by MaxU
use groupby(..., sort=False).sum():
使用groupby(..., sort=False).sum()：
In [84]: (testdf.reset_index()
....: .replace({'index': {'Uncoded & errors':'Other / unknown'}})
....: .groupby('index', sort=False).sum()
....: )
Out[84]:
counts percents
index
Daylight 10541 23.07
Dawn 4143 9.07
Other / unknown 754 1.65
Total 45690 100.00
回答by peterfields
Although I prefer MaxU's answer, you can also try summing in-place:
虽然我更喜欢 MaxU 的答案，但您也可以尝试就地求和：
testdf.loc['Other / unknown'] += testdf.loc['Uncoded & errors']
And then deleting the row by index:
然后按索引删除行：
testdf.drop(['Uncoded & errors'], inplace=True)
In [28]: testdf
Out[28]:
counts percents
Daylight 10541 23.07
Dawn 4143 9.07
Other / unknown 754 1.65
Total 45690 100.00
