我在pandas中有一个数据框,其组织如下:
btc_price['btc_price'] = pd.to_numeric(btc_price['btc_price'].str.replace(',', ''))
btc_price.head(n=120)
Out[4]:
btc_price
time
2017-08-27 22:50:00 4,389.6113
2017-08-27 22:51:00 4,389.0850
2017-08-27 22:52:00 4,388.8625
2017-08-27 22:53:00 4,389.7888
2017-08-27 22:56:00 4,389.9138
2017-08-27 22:57:00 4,390.1663
2017-08-27 22:58:00 4,390.2600
2017-08-27 22:59:00 4,392.4013
2017-08-27 23:00:00 4,391.6588
2017-08-27 23:01:00 4,391.9213
2017-08-27 23:02:00 4,394.0113
2017-08-27 23:03:00 4,396.9713
2017-08-27 23:04:00 4,397.3350
2017-08-27 23:05:00 4,397.0700
2017-08-27 23:06:00 4,398.6188
2017-08-27 23:07:00 4,398.5725
2017-08-27 23:08:00 4,397.4713
2017-08-27 23:09:00 4,398.0938
2017-08-27 23:10:00 4,398.7775
2017-08-27 23:11:00 4,398.0200
2017-08-27 23:12:00 4,397.9513
2017-08-27 23:13:00 4,398.0613
2017-08-27 23:14:00 4,398.0900
2017-08-27 23:15:00 4,398.0063
2017-08-27 23:16:00 4,397.6088
2017-08-27 23:17:00 4,394.3763
2017-08-27 23:46:00 4,389.1100
2017-08-27 23:48:00 4,390.6763
2017-08-27 23:49:00 4,392.5388
2017-08-27 23:49:00 4,392.5388
...
2017-08-28 00:51:00 4,367.5738
2017-08-28 00:51:00 4,367.5738
2017-08-28 00:52:00 4,367.7888
2017-08-28 00:53:00 4,368.4188
2017-08-28 00:54:00 4,368.8225
2017-08-28 00:55:00 4,368.7438
2017-08-28 00:57:00 4,368.4700
2017-08-28 00:58:00 4,367.9963
2017-08-28 00:59:00 4,366.4750
2017-08-28 01:00:00 4,359.1988
2017-08-28 01:01:00 4,355.2825
2017-08-28 01:02:00 4,352.3675
2017-08-28 01:03:00 4,354.2188
2017-08-28 01:04:00 4,353.5263
2017-08-28 01:05:00 4,354.2488
2017-08-28 01:06:00 4,358.8063
2017-08-28 01:07:00 4,359.5738
2017-08-28 01:08:00 4,361.7313
2017-08-28 01:09:00 4,360.8638
2017-08-28 01:10:00 4,363.0750
2017-08-28 01:11:00 4,362.3375
2017-08-28 01:12:00 4,362.3338
2017-08-28 01:13:00 4,358.8000
2017-08-28 01:14:00 4,354.0463
2017-08-28 01:15:00 4,356.1950
2017-08-28 01:16:00 4,359.5975
2017-08-28 01:17:00 4,360.1588
2017-08-28 01:18:00 4,362.2338
2017-08-28 01:19:00 4,363.7900
2017-08-28 01:20:00 4,362.6150
我想创建一个值为-1,0,1的新列change
。这些应该相当于过去一小时(-1)价格下降5%,"没有变化" (0),过去一小时(1)价格上涨5%。此外,一小时的价值应该是可变的,所以我可以将它改为一天或30分钟,例如,我觉得合适。
我发现了类似的问题here和here,但我是python的新手,并不确定如何将这些解决方案专门应用于我的问题。
另一个选择是计算每小时的平均价格,然后按小时计算百分比变化,但我希望能够使用滚动时间范围。
我也尝试在R中做这个没有运气。请帮忙。
我开始尝试:
btc_price['change'] = btc_price.pct_change(periods=60, fill_method='pad', limit=None, freq=None)
这有效,但并不能完全满足我的要求,我想将每个值与过去"时间框架的最小值和最大值进行比较"并根据此值计算%变化,而不是简单地比较两行。
我最终想要的是这样的(不完整的):
# Calculate the % change in btc_price
def calc_change(df):
array = df.values
a = array[:,1]
# Apply % change comparison to timeframe
def rolling(df, period, func, min_periods = None):
if min_periods is None:
min_periods = period
result = pd.Series(np.nan, index = df.index)
for i in range(1, len(df) + 1):
sub_df = df.iloc[max(i)]
我相信我可以使用类似df.rolling()
here之类的内容,但我不太确定这是否正是我想要的,因为我不太明白它是如何工作的。解释会很棒。
答案 0 :(得分:6)
pd.read_clipboard
pd.to_numeric
投射它。拥有有效数据后,您可以执行以下操作:
In [59]: df.head()
Out[59]:
btc_price
time
2017-09-07 22:50:00 4389.6113
2017-09-07 22:51:00 4389.0850
2017-09-07 22:52:00 4388.8625
2017-09-07 22:53:00 4389.7888
2017-09-07 22:56:00 4389.9138
In [60]: df = df.resample('1MIN').ffill(); df.head(10)
Out[60]:
btc_price
time
2017-09-07 22:50:00 4389.6113
2017-09-07 22:51:00 4389.0850
2017-09-07 22:52:00 4388.8625
2017-09-07 22:53:00 4389.7888
2017-09-07 22:54:00 4389.7888
2017-09-07 22:55:00 4389.7888
2017-09-07 22:56:00 4389.9138
2017-09-07 22:57:00 4390.1663
2017-09-07 22:58:00 4390.2600
2017-09-07 22:59:00 4392.4013
In [61]: WINDOW = 5 # 5 minutes, you can change to any window you want. Has to match resolution from resample
In [63]: df['change'] = df['btc_price'].pct_change(periods=WINDOW); df.head(10)
Out[63]:
btc_price change
time
2017-09-07 22:50:00 4389.6113 NaN
2017-09-07 22:51:00 4389.0850 NaN
2017-09-07 22:52:00 4388.8625 NaN
2017-09-07 22:53:00 4389.7888 NaN
2017-09-07 22:54:00 4389.7888 NaN
2017-09-07 22:55:00 4389.7888 0.000040
2017-09-07 22:56:00 4389.9138 0.000189
2017-09-07 22:57:00 4390.1663 0.000297
2017-09-07 22:58:00 4390.2600 0.000107
2017-09-07 22:59:00 4392.4013 0.000595
In [64]: import numpy as np
]n [67]: df['change_label'] = pd.cut(df['change'], [np.NINF, -0.05, 0.05, np.PINF], labels=['below 5%', 'around 0%', 'above 5%'])
In [69]: df.head(10)
Out[69]:
btc_price change change_label
time
2017-09-07 22:50:00 4389.6113 NaN NaN
2017-09-07 22:51:00 4389.0850 NaN NaN
2017-09-07 22:52:00 4388.8625 NaN NaN
2017-09-07 22:53:00 4389.7888 NaN NaN
2017-09-07 22:54:00 4389.7888 NaN NaN
2017-09-07 22:55:00 4389.7888 4.043638e-05 around 0%
2017-09-07 22:56:00 4389.9138 1.888321e-04 around 0%
2017-09-07 22:57:00 4390.1663 2.970701e-04 around 0%
2017-09-07 22:58:00 4390.2600 1.073400e-04 around 0%
2017-09-07 22:59:00 4392.4013 5.951311e-04 around 0%
感觉你需要:
Resample
以获得可预测的解决方案FFill
为了没有洞。或者在你的情况下以其他方式处理它。pct_change
。pd.cut
。此外,简单的df['change'].map(lamba v: # here logic)
也可以。