在 Pandas 中减少一列
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Reductions down a column in Pandas
提问by Isaac
I'm trying to transform a (well, many) column of return data to a column of closing prices. In Clojure, I'd use reductions, which is like reduce, but returns a sequence of all the intermediate values.
我正在尝试将一列（很多）返回数据转换为一列收盘价。在 Clojure 中，我会使用reductions，它类似于reduce，但返回所有中间值的序列。
e.g.
例如
$ c
0.12
-.13
0.23
0.17
0.29
-0.11
# something like this
$ c.reductions(init=1, lambda accumulator, ret: accumulator * (1 + ret))
1.12
0.97
1.20
1.40
1.81
1.61
NB: The actual closing price doesn't matter, hence using 1 as the initial value. I just need a "mock" closing price.
注意：实际收盘价无关紧要，因此使用 1 作为初始值。我只需要一个“模拟”收盘价。
My data's actual structure is a DataFrame of named columns of TimeSeries. I guess I'm looking for a function similar applymap, but I'd rather not do something hacky with that function and reference the DF from within it (which I suppose is one solution to this problem?)
我的数据的实际结构是 TimeSeries 命名列的 DataFrame。我想我正在寻找一个类似的函数applymap，但我宁愿不对该函数做一些hacky 并从其中引用 DF（我认为这是解决此问题的一种方法？）
Additionally, what would I do if I wanted to keep the returnsdata, but have the closing "price" with it? Should I return a tuple instead, and have the TimeSeries be of the type (returns, closing_price)?
此外，如果我想保留returns数据，但有收盘价，我该怎么办？我应该返回一个元组，并将 TimeSeries 设为类型(returns, closing_price)吗？
采纳答案by Andy Hayden
It's worth noting that it's often faster (as well as easier to understand) to write more verbosely in pandas, rather than write as a reduce.
值得注意的是，在 Pandas 中更详细地编写通常更快（也更容易理解），而不是编写为reduce.
In your specific example I would just addand then cumprod:
In [2]: c.add(1).cumprod()
Out[2]:
0 1.120000
1 0.974400
2 1.198512
3 1.402259
4 1.808914
5 1.609934
or perhaps init * c.add(1).cumprod().
或者也许init * c.add(1).cumprod()。
Note: In some cases however, for example where memory is an issue, you may have to rewrite these in a more low-level/clever way, but it's usually worth trying the simplest method first (and testing against it e.g. using %timeit or profiling memory).
注意：然而，在某些情况下，例如在内存有问题的情况下，您可能必须以更低级/更聪明的方式重写它们，但通常值得先尝试最简单的方法（并针对它进行测试，例如使用 %timeit 或分析内存）。
回答by Zelazny7
It doesn't look like it's a well publicized feature yet, but you can use expanding_applyto achieve the returns calculation:
它看起来还不是一个广为人知的功能，但您可以使用它expanding_apply来实现收益计算：
In [1]: s
Out[1]:
0 0.12
1 -0.13
2 0.23
3 0.17
4 0.29
5 -0.11
In [2]: pd.expanding_apply(s ,lambda s: reduce(lambda x, y: x * (1+y), s, 1))
Out[2]:
0 1.120000
1 0.974400
2 1.198512
3 1.402259
4 1.808914
5 1.609934
I'm not 100% certain, but I believe expanding_applyworks on the applied series starting from the first index through the current index. I use the built-in reducefunction that works exactly like your Clojure function.
我不是 100% 确定，但我相信expanding_apply适用于从第一个索引到当前索引的应用系列。我使用的内置reduce函数与您的 Clojure 函数完全一样。
Docstring for expanding_apply:
文档字符串expanding_apply：
Generic expanding function application
Parameters
----------
arg : Series, DataFrame
func : function
Must produce a single value from an ndarray input
min_periods : int
Minimum number of observations in window required to have a value
freq : None or string alias / date offset object, default=None
Frequency to conform to before computing statistic
center : boolean, default False
Whether the label should correspond with center of window
Returns
-------
y : type of input argument
回答by Alexander
For readability, I prefer the following solution:
为了可读性，我更喜欢以下解决方案：
returns = pd.Series([0.12, -.13, 0.23, 0.17, 0.29, -0.11])
initial_value = 100
cum_growth = initial_value * (1 + returns).cumprod()
>>> cum_growth
0 112.000000
1 97.440000
2 119.851200
3 140.225904
4 180.891416
5 160.993360
dtype: float64
If you'd like to include the initial value in the series:
如果您想在系列中包含初始值：
>>> pd.concat([pd.Series(initial_value), cum_growth]).reset_index(drop=True)
0 100.000000
1 112.000000
2 97.440000
3 119.851200
4 140.225904
5 180.891416
6 160.993360
dtype: float64
