我正在尝试以1分钟为基础的数据帧计算4小时的回报。
frequency = int(1*60*4)
a = data[1:]['close'] / data[:-frequency]['close'].values - 1
但是我收到了这个错误：
ValueError: operands could not be broadcast together with shapes (253062,) (252823,)
当我按分钟返回时也这样做：
a = data[1:]['close'] / data[:-1]['close'].values - 1
我该如何解决这个问题？
这是样本ata：
date open high low close t
0 20150101 130000 1.20965 1.20977 1.20962 1.20962 0
1 20150101 130100 1.20963 1.20968 1.20962 1.20962 0
2 20150101 130200 1.20965 1.20970 1.20961 1.20961 0
3 20150101 130400 1.20959 1.21008 1.20959 1.20983 0
4 20150101 130500 1.20988 1.20988 1.20988 1.20988 0
5 20150101 130600 1.20984 1.20984 1.20982 1.20982 0
6 20150101 130700 1.20986 1.20999 1.20986 1.20987 0
7 20150101 130800 1.20998 1.21022 1.20987 1.21008 0
8 20150101 130900 1.20996 1.20996 1.20996 1.20996 0
9 20150101 131000 1.21013 1.21019 1.20967 1.20977 0
10 20150101 131100 1.20976 1.20999 1.20976 1.20988 0
11 20150101 131200 1.20987 1.20987 1.20987 1.20987 0
12 20150101 131300 1.20997 1.21006 1.20993 1.21006 0
13 20150101 131400 1.21007 1.21007 1.21006 1.21006 0
14 20150101 131600 1.21004 1.21004 1.21004 1.21004 0
15 20150101 131800 1.21003 1.21004 1.20979 1.20979 0
16 20150101 132700 1.21003 1.21003 1.20979 1.21003 0
17 20150101 132800 1.20979 1.21003 1.20979 1.21003 0
18 20150101 132900 1.21001 1.21003 1.20999 1.21003 0
19 20150101 133100 1.21033 1.21041 1.21033 1.21033 0
20 20150101 133200 1.21028 1.21035 1.21006 1.21035 0
21 20150101 133500 1.21005 1.21006 1.21005 1.21005 0
22 20150101 133600 1.21004 1.21006 1.21004 1.21006 0
23 20150101 133700 1.20991 1.21006 1.20991 1.21004 0
24 20150101 133800 1.21003 1.21004 1.20982 1.21004 0
25 20150101 133900 1.21019 1.21034 1.21019 1.21034 0
26 20150101 134000 1.21030 1.21034 1.21030 1.21032 0
27 20150101 134100 1.21006 1.21007 1.21006 1.21007 0
28 20150101 134300 1.21003 1.21006 1.21003 1.21006 0
29 20150101 134400 1.21003 1.21007 1.21003 1.21007 0
... ... ... ... ... ... ..
253033 20150904 162900 1.11511 1.11515 1.11499 1.11509 0
253034 20150904 163000 1.11509 1.11511 1.11507 1.11507 0
253035 20150904 163100 1.11507 1.11530 1.11507 1.11524 0
253036 20150904 163200 1.11521 1.11546 1.11520 1.11537 0
253037 20150904 163300 1.11533 1.11533 1.11520 1.11528 0
253038 20150904 163400 1.11528 1.11528 1.11528 1.11528 0
253039 20150904 163500 1.11527 1.11527 1.11486 1.11491 0
253040 20150904 163600 1.11492 1.11517 1.11489 1.11513 0
253041 20150904 163700 1.11513 1.11513 1.11499 1.11503 0
253042 20150904 163800 1.11498 1.11502 1.11482 1.11491 0
253043 20150904 163900 1.11490 1.11491 1.11489 1.11491 0
253044 20150904 164000 1.11490 1.11491 1.11490 1.11490 0
253045 20150904 164100 1.11488 1.11488 1.11477 1.11480 0
253046 20150904 164200 1.11482 1.11483 1.11481 1.11483 0
253047 20150904 164300 1.11482 1.11484 1.11482 1.11483 0
253048 20150904 164400 1.11483 1.11484 1.11480 1.11484 0
253049 20150904 164500 1.11480 1.11502 1.11480 1.11501 0
253050 20150904 164600 1.11502 1.11506 1.11488 1.11506 0
253051 20150904 164700 1.11501 1.11501 1.11496 1.11501 0
253052 20150904 164800 1.11501 1.11510 1.11499 1.11505 0
253053 20150904 164900 1.11504 1.11520 1.11503 1.11504 0
253054 20150904 165000 1.11506 1.11513 1.11502 1.11511 0
253055 20150904 165100 1.11509 1.11509 1.11500 1.11501 0
253056 20150904 165200 1.11500 1.11510 1.11500 1.11500 0
253057 20150904 165300 1.11516 1.11516 1.11498 1.11505 0
253058 20150904 165400 1.11503 1.11503 1.11454 1.11469 0
253059 20150904 165500 1.11472 1.11472 1.11454 1.11462 0
253060 20150904 165600 1.11462 1.11487 1.11447 1.11479 0
253061 20150904 165700 1.11484 1.11506 1.11477 1.11497 0
253062 20150904 165800 1.11495 1.11497 1.11432 1.11435 0
答案 0 :(得分：2)
而不是自己安排切片，
您可以使用remaining_balance = (annualInterestRate/12 * (balance-(balance * monthlyPaymentRate)) + tem_balance)方法计算变化百分比。
如果您将日期字符串转换为Pandas时间戳并将pct_change列设置为索引，那么您可以使用参数date指定4小时频率：
freq='4H' data['date'] = pd.to_datetime(data['date'], format='%Y%m%d %H%M%S')
data = data.set_index('date')
a = data['close'].pct_change(freq='4H')
是一个长度为252823的数组。
data[:-frequency]['close'].values是一系列长度253062。
要彼此分开（元素方向），两者必须具有相同的长度。
因此，要在没有data[1:]['close']的情况下进行计算，您需要使用
pct_change例如，
a = data[frequency:]['close'] / data[:-frequency]['close'].values - 1
比较In [182]: N = 7; s = pd.Series(range(N), index=pd.date_range('2000-1-1', periods=N, freq='H')); s
Out[182]:
2000-01-01 00:00:00 0
2000-01-01 01:00:00 1
2000-01-01 02:00:00 2
2000-01-01 03:00:00 3
2000-01-01 04:00:00 4
2000-01-01 05:00:00 5
2000-01-01 06:00:00 6
Freq: H, dtype: int64
：
pct_change与分裂的结果：
In [183]: s.pct_change(freq='4H')
Out[183]:
2000-01-01 00:00:00 NaN
2000-01-01 01:00:00 NaN
2000-01-01 02:00:00 NaN
2000-01-01 03:00:00 NaN
2000-01-01 04:00:00 inf
2000-01-01 05:00:00 4.000000
2000-01-01 06:00:00 2.000000
2000-01-01 07:00:00 NaN
2000-01-01 08:00:00 NaN
2000-01-01 09:00:00 NaN
2000-01-01 10:00:00 NaN
Freq: H, dtype: float64