,所以我有一个数据集,如下:
# Example
0 1 2 3 4 5
0 18 1 -19 -16 -5 19
1 18 0 -19 -17 -6 19
2 17 -1 -20 -17 -6 19
3 18 1 -19 -16 -5 20
4 18 0 -19 -16 -5 20
实际数据:
[{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 19},
{0: 18, 1: 0, 2: -19, 3: -17, 4: -6, 5: 19},
{0: 17, 1: -1, 2: -20, 3: -17, 4: -6, 5: 19},
{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -20, 3: -15, 4: -4, 5: 20},
{0: 19, 1: 1, 2: -18, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -17, 4: -7, 5: 18},
{0: 18, 1: 0, 2: -20, 3: -18, 4: -7, 5: 18},
{0: 17, 1: 0, 2: -19, 3: -17, 4: -7, 5: 18},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 1, 2: -18, 3: -16, 4: -5, 5: 20},
{0: 17, 1: 0, 2: -20, 3: -16, 4: -5, 5: 19},
{0: 17, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -15, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -14, 4: -3, 5: 22},
{0: 18, 1: 1, 2: -18, 3: -14, 4: -4, 5: 22}]
上述形状将是: (20, 6)
。
我要实现的是在当时4行上的每一列应用自定义功能。
示例:
- 第一次迭代 ->
f()
用于所有列df.ix[0:3]
; - 第二次迭代 ->
f()
用于所有列df.ix[4:7]
;
等等...
在某种程度上,我需要的是大小4的滚动窗口4。
结果使用上述数据时,将是形状的数据框架:(5, 6)
。只是为了参数,您可以假设自定义功能是每列的4行的平均值。
到目前为止我尝试了什么?
- 我正在考虑滚动,但滚动并不做我需要做的事情。它滚动一个窗户,大步为1。
- 实际实施了它,但是由于数据量: ,我确实需要对其进行优化。
这是代码:
curr = 0
res = []
while curr < df_to_look_at2.shape[0]:
look_at = df_to_look_at2.ix[curr:curr+3]
curr += 4
res.append(look_at.mean().values.tolist())
pd.DataFrame(res)
和结果:
0 1 2 3 4 5
0 17.75 0.25 -19.25 -16.50 -5.50 19.25
1 18.25 0.25 -19.00 -16.00 -5.25 19.50
2 17.75 0.25 -19.25 -16.75 -5.75 19.00
3 17.75 0.25 -19.00 -16.00 -4.75 19.75
4 17.75 0.25 -18.75 -14.75 -3.75 21.00
一个额外的想法,如果它不仅含义是均值,而是min((,max((,shay((和其他一些自定义函数...
如果您想在一个以上的窗口中考虑不止一次,则在这里滚动将是准确的。但是,您的窗户是唯一的,因此您真正问的是如何通过arange
和地板部门进行分组。
window_size = 4
grouper = np.arange(df.shape[0]) // window_size
df.groupby(grouper).mean()
0 1 2 3 4 5
0 17.75 0.25 -19.25 -16.50 -5.50 19.25
1 18.25 0.25 -19.00 -16.00 -5.25 19.50
2 17.75 0.25 -19.25 -16.75 -5.75 19.00
3 17.75 0.25 -19.00 -16.00 -4.75 19.75
4 17.75 0.25 -18.75 -14.75 -3.75 21.00
我认为以这种方式的多个计算确实属于Numpy Turf。您可以使用Reshape以所需的格式获取基础数组,然后根据需要在数组上计算。
inp = [{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 19},
{0: 18, 1: 0, 2: -19, 3: -17, 4: -6, 5: 19},
{0: 17, 1: -1, 2: -20, 3: -17, 4: -6, 5: 19},
{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -20, 3: -15, 4: -4, 5: 20},
{0: 19, 1: 1, 2: -18, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -17, 4: -7, 5: 18},
{0: 18, 1: 0, 2: -20, 3: -18, 4: -7, 5: 18},
{0: 17, 1: 0, 2: -19, 3: -17, 4: -7, 5: 18},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 1, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -16, 4: -5, 5: 20},
{0: 18, 1: 1, 2: -18, 3: -16, 4: -5, 5: 20},
{0: 17, 1: 0, 2: -20, 3: -16, 4: -5, 5: 19},
{0: 17, 1: 0, 2: -19, 3: -16, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -15, 4: -4, 5: 20},
{0: 18, 1: 0, 2: -19, 3: -14, 4: -3, 5: 22},
{0: 18, 1: 1, 2: -18, 3: -14, 4: -4, 5: 22}]
import pandas as pd
df = pd.DataFrame(inp)
temp = df.values.reshape(-1, 4, df.shape[-1])
out = pd.DataFrame(temp.mean(axis=1))
输出:
0 1 2 3 4 5
0 17.75 0.25 -19.25 -16.50 -5.50 19.25
1 18.25 0.25 -19.00 -16.00 -5.25 19.50
2 17.75 0.25 -19.25 -16.75 -5.75 19.00
3 17.75 0.25 -19.00 -16.00 -4.75 19.75
4 17.75 0.25 -18.75 -14.75 -3.75 21.00