前 x 值的平均值答案

【问题标题】：Mean of the first x values前 x 值的平均值
【发布时间】：2020-10-23 07:02:35
【问题描述】：

假设我有一张如下表。每行代表一个群组，每列代表一个时间段。有些值是 nan，因为该时间段尚未发生。

对于每一列，我想获得前四个值的平均值（不包括 nan）。有没有简单的方法可以在 pandas 中实现这一点？

| name     |   1 |          2 |           3 |          4 |            5 |            6 |           7 |            8 |            9 |           10 |           11 |           12 |
|----------|----:|-----------:|------------:|-----------:|-------------:|-------------:|------------:|-------------:|-------------:|-------------:|-------------:|-------------:|
| 2019     | nan | -0.0127273 | -0.0202578  | -1         | nan          | nan          | nan         | nan          | nan          | nan          | nan          | nan          |
| 2018     | nan | -0.0245184 | -0.0143627  | -0.0309654 |  -0.0131579  |  -0.0171429  |  -1         | nan          | nan          | nan          | nan          | nan          |
| 2017     | nan | -0.0122807 | -0.00888099 | -0.0340502 |  -0.0222635  |  -0.0151803  |  -0.0346821 |   0.011976   |  -0.0118343  |  -1          | nan          | nan          |
| 2016     | nan | -0.0196429 | -0.010929   | -0.0570902 |  -0.0195312  |   0.00199203 |  -0.027833  |  -0.0102249  |  -0.0103306  |   0.00835073 |  -0.00828157 |  -0.0605428  |
| 2015     | nan | -0.0158451 | -0.0125224  | -0.0525362 |  -0.0114723  |  -0.0154739  |  -0.0314342 |  -0.00811359 |  -0.00817996 |  -0.00206186 |   0.00206612 |  -0.0206186  |
| 2014     | nan | -0.0175439 | -0.00892857 | -0.0486486 |  -0.00757576 |  -0.0114504  |  -0.030888  |  -0.0059761  |  -0.0140281  |   0.0121951  |  -0.00803213 |  -0.0425101  |
| 2013     | nan | -0.0250447 | -0.00550459 | -0.0682657 |  -0.0158416  |  -0.00402414 |  -0.0323232 |  -0.0167015  |  -0.00212314 |  -0.0191489  |   0.0021692  |  -0.017316   |
| 2012     | nan | -0.0118443 | -0.00171233 | -0.0703259 |  -0.00184502 |   0.00554529 |  -0.0294118 |  -0.00757576 |  -0.00763359 |  -0.00769231 |  -0.00387597 |   0.00194553 |
| 2011     | nan | -0.0174825 | -0.0106762  | -0.0377698 |  -0.0261682  |  -0.00767754 |  -0.0174081 |  -0.015748   |  -0.004      |  -0.0200803  |   0.00409836 |  -0.0285714  |
| 2010     | nan | -0.0234657 | -0.00369686 | -0.0352505 |  -0.0269231  |   0.00395257 |  -0.019685  |  -0.0100402  |  -0.00608519 |  -0.0102041  |  -0.0123711  |  -0.00626305 |

【问题讨论】：

标签： pandas

【解决方案1】：

IIUC，我们可以通过堆栈来移除任何 NaN 值，然后应用 groupby 来获取每个值的计数，从而创建排序的伪索引。

然后我们可以传递一个最终的 groupby 并将平均值连接到底部。

s = df.iloc[:, 1:].T.stack().to_frame("vals")
s["idx"] = s.groupby(level=0).cumcount()
s1 = (
    s[s["idx"].le(3)]
    .groupby(level=0)
    .mean()
    .drop("idx", 1)
    .stack()
    .unstack(0)
    .reset_index(0, drop=True)
)
# s1.columns = s1.columns.astype(str) if columns are strings.

df_mean = pd.concat([df,s1])

另一种方法是用 append 融化。

s = pd.melt(df.iloc[:,1:]).dropna()
s1 = s.assign(key=s.groupby("variable").cumcount()).query("key <= 3").groupby("variable")[
    "value"
].mean().to_frame('mean').T
df_mean = df.append(s1)

        name   1         2         3         4         5         6         7  \
1     2019.0 NaN -0.012727 -0.020258 -1.000000       NaN       NaN       NaN   
2     2018.0 NaN -0.024518 -0.014363 -0.030965 -0.013158 -0.017143 -1.000000   
3     2017.0 NaN -0.012281 -0.008881 -0.034050 -0.022263 -0.015180 -0.034682   
4     2016.0 NaN -0.019643 -0.010929 -0.057090 -0.019531  0.001992 -0.027833   
5     2015.0 NaN -0.015845 -0.012522 -0.052536 -0.011472 -0.015474 -0.031434   
6     2014.0 NaN -0.017544 -0.008929 -0.048649 -0.007576 -0.011450 -0.030888   
7     2013.0 NaN -0.025045 -0.005505 -0.068266 -0.015842 -0.004024 -0.032323   
8     2012.0 NaN -0.011844 -0.001712 -0.070326 -0.001845  0.005545 -0.029412   
9     2011.0 NaN -0.017483 -0.010676 -0.037770 -0.026168 -0.007678 -0.017408   
10    2010.0 NaN -0.023466 -0.003697 -0.035250 -0.026923  0.003953 -0.019685   
mean     NaN NaN -0.017292 -0.013608 -0.280526 -0.016606 -0.011451 -0.273487   

             8         9        10        11        12  
1          NaN       NaN       NaN       NaN       NaN  
2          NaN       NaN       NaN       NaN       NaN  
3     0.011976 -0.011834 -1.000000       NaN       NaN  
4    -0.010225 -0.010331  0.008351 -0.008282 -0.060543  
5    -0.008114 -0.008180 -0.002062  0.002066 -0.020619  
6    -0.005976 -0.014028  0.012195 -0.008032 -0.042510  
7    -0.016702 -0.002123 -0.019149  0.002169 -0.017316  
8    -0.007576 -0.007634 -0.007692 -0.003876  0.001946  
9    -0.015748 -0.004000 -0.020080  0.004098 -0.028571  
10   -0.010040 -0.006085 -0.010204 -0.012371 -0.006263  
mean -0.003085 -0.011093 -0.245379 -0.003020 -0.035247

【讨论】：