Pandas：累积回报函数
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Pandas: Cumulative return function
提问by Kelaref
I have a dataframe such as the following:
我有一个数据框，如下所示：
Index Return
2008-11-21 0.153419
2008-11-24 0.037421
2008-11-25 0.077500
What's the best way to calculate a cumulative return across all columns on the last row?
计算最后一行所有列的累积回报的最佳方法是什么？
Following is the intended result:
以下是预期的结果：
Index Return
2008-11-21 0.153419
2008-11-24 0.037421
2008-11-25 0.077500
Cumulative 0.289316
Where cumulative return calculated as follows:
其中累积回报计算如下：
cumulative = (1 + return1) * (1 + return2) * (1 + return3) - 1
What is the best way to perform this in pandas?
在Pandas中执行此操作的最佳方法是什么？
回答by Steven G
there is a pandas cumprod()method for that. this will work for every columns.
有一个Pandascumprod()方法。这将适用于每一列。
df.ix["Cumulative"] = ((df+1).cumprod()-1).iloc[-1]
this would be about 2 time faster than other solutions on large dataset:
这将比大型数据集上的其他解决方案快约 2 倍：
In[106]: %timeit df.ix["Cumulative"] = ((df+1).cumprod()-1).iloc[-1]
10 loops, best of 3: 18.4 ms per loop
In[107]: %timeit df.ix['Cummulative'] = df.apply(lambda x: (x+1).prod()-1)
10 loops, best of 3: 32.9 ms per loop
In[110]: %timeit df.append(df.iloc[:,1:].apply(lambda col: (col + 1).prod() - 1), ignore_index=True)
10 loops, best of 3: 37.1 ms per loop
In[113]: %timeit df.append(df.apply(lambda col: prod([(1+c) for c in col]) - 1), ignore_index=True)
1 loop, best of 3: 262 ms per loop
I would suggest to neveruse apply if you can find a built-in method since apply is looping over the dataframe which makes it slow. Bult-in method are highly efficient and normally there is no way you are going to get faster than them using apply.
如果您能找到内置方法，我建议永远不要使用 apply，因为 apply 正在循环遍历数据帧，这使得它变慢。内置方法非常有效，通常没有办法比使用 apply 更快。
回答by TheF1rstPancake
Another solution:
另一种解决方案：
df.ix["Cumulative"] = (df['Return']+1).prod() - 1
This will add 1 to the df['Return']column, multiply all the rows together, and then subtract one from the result. This will result in a simple float value. The result will then be placed at the index "Cumulative". Since that index doesn't exist yet, it will be appended to the end of the DataFrame:
这会将df['Return']列加 1 ，将所有行相乘，然后从结果中减去 1。这将产生一个简单的浮点值。结果将被放置在索引“累积”中。由于该索引尚不存在，它将被附加到 DataFrame 的末尾：
Return
2008-11-21 0.153419
2008-11-25 0.077500
2008-11-24 0.037421
Cummulative 0.289316
If you want to apply this across multiple columns:
如果您想在多列中应用它：
df.ix['Cummulative'] = df.apply(lambda x: (x+1).prod()-1)
This would output the following (I made a second column called "Return2" that is a copy of "Return"):
这将输出以下内容（我创建了名为“Return2”的第二列，它是“Return”的副本）：
Return Return2
2008-11-21 0.153419 0.153419
2008-11-25 0.077500 0.077500
2008-11-24 0.037421 0.037421
Cummulative 0.289316 0.289316
回答by Psidom
With pandas, you can use the prod()method:
使用pandas，您可以使用以下prod()方法：
df.append(df.iloc[:,1:].apply(lambda col: (col + 1).prod() - 1), ignore_index=True)
# Index Return
#0 2008-11-21 0.153419
#1 2008-11-24 0.037421
#2 2008-11-25 0.077500
#3 NaN 0.289316
Or as @Randy C commented, this can be further simplified to:
或者正如@Randy C 所评论的，这可以进一步简化为：
df.append((df.iloc[:,1:] + 1).prod() - 1, ignore_index=True)
回答by AlexG
Here is mine:
这是我的：
from numpy import prod
df.append(df.apply(lambda col: prod([(1+c) for c in col]) - 1), ignore_index=True)
回答by Randy
One option is to just use reduce, though others might be able to come up with faster vectorized methods:
一种选择是仅使用reduce，尽管其他人可能能够提出更快的矢量化方法：
In [10]: pd.read_clipboard()
Out[10]:
Index Return
0 2008-11-21 0.153419
1 2008-11-24 0.037421
2 2008-11-25 0.077500
In [11]: reduce(lambda x, y: (1+x)*(1+y)-1, _10['Return'])
Out[11]: 0.28931612705992227
Note that in Python 3, reduceis part of the functoolslibrary, though it's a builtin for Python 2.
请注意，在 Python 3 中，它reduce是functools库的一部分，尽管它是 Python 2 的内置函数。
