我们说我有一个看起来像这样的DataFrame：
In [41]: df.columns
Out[41]: Index([u'Date Time', u'Open', u'High', u'Low', u'Last'], dtype='object')
In [42]: df
Out[42]:
Date Time Open High Low Last
0 12/02/2007 23:23:00 1443.75 1444.00 1443.75 1444.00
1 12/02/2007 23:25:00 1444.00 1444.00 1444.00 1444.00
2 12/02/2007 23:26:00 1444.25 1444.25 1444.25 1444.25
3 12/02/2007 23:27:00 1444.25 1444.25 1444.25 1444.25
4 12/02/2007 23:28:00 1444.25 1444.25 1444.25 1444.25
5 12/02/2007 23:29:00 1444.25 1444.25 1444.00 1444.00
6 12/02/2007 23:30:00 1444.25 1444.25 1444.00 1444.00
7 12/02/2007 23:31:00 1444.25 1444.25 1443.75 1444.00
8 12/02/2007 23:32:00 1444.00 1444.00 1443.75 1443.75
9 12/02/2007 23:33:00 1444.00 1444.00 1443.50 1443.50
我想创建一个关联＆＃39;日期时间＆＃39;当前索引的列，包含此列和前面n个索引的其余列。例如，当index = 9和n = 2时，目标结果将转换这些行：
7 12/02/2007 23:31:00 1444.25 1444.25 1443.75 1444.00
8 12/02/2007 23:32:00 1444.00 1444.00 1443.75 1443.75
9 12/02/2007 23:33:00 1444.00 1444.00 1443.50 1443.50
进入具有以下值的列表，其中索引1-4来自第9行，5-8来自第8行，9-12来自第7行：
['12/02/2007 23:33:00', 1444.00, 1444.00, 1443.50, 1443.50, 1444.00, 1444.00, 1443.75, 1443.75, 1444.25, 1444.25, 1443.75, 1444.00]
我确信我可以轻松地迭代数据帧的片段并创建数组，但我希望有一种更有效的方法。
编辑：
以下是一些生成我正在寻找的结果的代码。一些响应表明我可能会查看rolling_apply或rolling_window函数，但我无法弄清楚它是如何工作的。
import pandas as pd
import numpy as np
data = pd.DataFrame([
['12/02/2007 23:23:00', 1443.75, 1444.00, 1443.75, 1444.00],
['12/02/2007 23:25:00', 1444.00, 1444.00, 1444.00, 1444.00],
['12/02/2007 23:26:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:27:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:28:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:29:00', 1444.25, 1444.25, 1444.00, 1444.00],
['12/02/2007 23:30:00', 1444.25, 1444.25, 1444.00, 1444.00],
['12/02/2007 23:31:00', 1444.25, 1444.25, 1443.75, 1444.00],
['12/02/2007 23:32:00', 1444.00, 1444.00, 1443.75, 1443.75],
['12/02/2007 23:33:00', 1444.00, 1444.00, 1443.50, 1443.50]
])
window_size = 6
# Prime the DataFrame using the date as the index
result = pd.DataFrame(
[data.iloc[0:window_size, 1:].values.flatten()],
[data.iloc[window_size - 1, 0]])
for t in data.iloc[window_size:, 1:].itertuples(index=True):
# drop the oldest values and append the new ones
new_features = result.tail(1).iloc[:, 4:].values.flatten()
new_features = np.append(new_features, list(t[1:]), 0)
# turn it into a DataFrame and append it to the ongoing result
new_df = pd.DataFrame([new_features], [t[0]])
result = result.append(new_df)
这种方法不是很快，所以我仍然对如何改进它感兴趣。
答案 0 :(得分：0)
这个简单的功能对我有用
import itertools
def collapse(df, index_loc, number):
return list(itertools.chain(*[list(df.loc[x].values) for x in xrange(index_loc - number, index_loc + 1)]))
其中df是你的数据框，index_loc是起始索引（假设你在示例中有整数索引），数字是你的＆＃39; n＆＃39;。只需使用values方法获取每个索引点的数据框值，然后将列表链接在一起....
答案 1 :(得分：0)
以下是一些生成我正在寻找的结果的代码。一些响应表明我可能会查看rolling_apply或rolling_window函数，但我无法弄清楚它是如何工作的。
import pandas as pd
import numpy as np
data = pd.DataFrame([
['12/02/2007 23:23:00', 1443.75, 1444.00, 1443.75, 1444.00],
['12/02/2007 23:25:00', 1444.00, 1444.00, 1444.00, 1444.00],
['12/02/2007 23:26:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:27:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:28:00', 1444.25, 1444.25, 1444.25, 1444.25],
['12/02/2007 23:29:00', 1444.25, 1444.25, 1444.00, 1444.00],
['12/02/2007 23:30:00', 1444.25, 1444.25, 1444.00, 1444.00],
['12/02/2007 23:31:00', 1444.25, 1444.25, 1443.75, 1444.00],
['12/02/2007 23:32:00', 1444.00, 1444.00, 1443.75, 1443.75],
['12/02/2007 23:33:00', 1444.00, 1444.00, 1443.50, 1443.50]
])
window_size = 6
# Prime the DataFrame using the date as the index
result = pd.DataFrame(
[data.iloc[0:window_size, 1:].values.flatten()],
[data.iloc[window_size - 1, 0]])
for t in data.iloc[window_size:, 1:].itertuples(index=True):
# drop the oldest values and append the new ones
new_features = result.tail(1).iloc[:, 4:].values.flatten()
new_features = np.append(new_features, list(t[1:]), 0)
# turn it into a DataFrame and append it to the ongoing result
new_df = pd.DataFrame([new_features], [t[0]])
result = result.append(new_df)
这可能不是非常有效，但它解决了这个问题。