修改代码以捕获大于-而不是完全匹配的值 [英] Modify code to capture values greater than - instead of exact match
问题描述
下面的代码可以很好地识别在接下来的行中值是命中还是未命中，并在输出列中显示满足条件的时间.
The following code works well for identifying whether a value is hit or missed over following rows and giving the output column showing the time the condition was met.
import datetime,numpy as np,pandas as pd;
nan = np.nan;
a = pd.DataFrame( {'price': {datetime.time(9, 0): 1, datetime.time(10, 0): 0, datetime.time(11, 0): 3, datetime.time(12, 0): 4, datetime.time(13, 0): 7, datetime.time(14, 0): 6, datetime.time(15, 0): 5, datetime.time(16, 0): 4, datetime.time(17, 0): 0, datetime.time(18, 0): 2, datetime.time(19, 0): 4, datetime.time(20, 0): 7}, 'reversal': {datetime.time(9, 0): nan, datetime.time(10, 0): nan, datetime.time(11, 0): nan, datetime.time(12, 0): nan, datetime.time(13, 0): nan,
datetime.time(14, 0): 6.0, datetime.time(15, 0): nan, datetime.time(16, 0): nan, datetime.time(17, 0): nan, datetime.time(18, 0): nan, datetime.time(19, 0): nan, datetime.time(20, 0): nan}});
a['target_hit_time']=a['target_miss_time']=nan;
a['target1']=a['reversal']+1;
a['target2']=a['reversal']-a['reversal'];
a.sort_index(1,inplace=True);
hits = a.ix[:,:-2].dropna();
for row,hit in hits.iterrows():
forwardRows = [row]<a['price'].index.values
targetHit = a.index.values[(hit['target1']==a['price'].values) & forwardRows][0];
targetMiss = a.index.values[(hit['target2']==a['price'].values) & forwardRows][0];
if targetHit>targetMiss:
a.loc[row,"target_miss_time"] = targetMiss;
else:
a.loc[row,"target_hit_time"] = targetHit;
a
此图显示了上述代码的输出，可以通过运行以下代码轻松地复制它们:
This image shows the output from the above code which can easily be reproduced by running this code:
我遇到的问题是，当将此代码用于实际数据时，价格可能不完全匹配和/或可能会有一定的差距.因此，如果我们看下面的图片:
The issue I have is that when this code is utilised on real data the price may not exactly match and/or may gap though a value. So if we look at the following image:
我们看到，如果我们正在寻找值>= 7.5而不仅仅是寻找值7.5，那么将满足target1标准.有人可以帮忙修改代码来实现这一点吗?
We see that target1 criteria would be met if we were looking for a value >= 7.5 and not just looking for the value 7.5. Can anybody help modify the code to achieve this please?
推荐答案
某些if和多数民众赞成全部:D ...
Some ifs and thats all :D...
import datetime,numpy as np,pandas as pd;
nan = np.nan;
a = pd.DataFrame( {'price': {datetime.time(9, 0): 1, datetime.time(10, 0): 0, datetime.time(11, 0): 3, datetime.time(12, 0): 4, datetime.time(13, 0): 7, datetime.time(14, 0): 6, datetime.time(15, 0): 5, datetime.time(16, 0): 4, datetime.time(17, 0): 2, datetime.time(18, 0): 2, datetime.time(19, 0): 4, datetime.time(20, 0): 8}, 'reversal': {datetime.time(9, 0): nan, datetime.time(10, 0): nan, datetime.time(11, 0): nan, datetime.time(12, 0): nan, datetime.time(13, 0): nan,
datetime.time(14, 0): 6.0, datetime.time(15, 0): nan, datetime.time(16, 0): nan, datetime.time(17, 0): nan, datetime.time(18, 0): nan, datetime.time(19, 0): nan, datetime.time(20, 0): nan}});
a['target_hit_time']=a['target_miss_time']=nan;
a['target1']=a['reversal']+1;
a['target2']=a['reversal']-a['reversal'];
a.sort_index(1,inplace=True);
hits = a.ix[:,:-2].dropna();
for row,hit in hits.iterrows():
forwardRows = a[a.index.values > row];
targetHit = hit['target1']<=forwardRows['price'].values;
targetMiss = hit['target2']==forwardRows['price'].values;
targetHit = forwardRows[targetHit].head(1).index.values;
targetMiss = forwardRows[targetMiss].head(1).index.values;
targetHit, targetMiss = \
targetHit[0] if targetHit else [], \
targetMiss[0] if targetMiss else [];
goMiss,goHit = False,False
if targetHit and targetMiss:
if targetHit>targetMiss: goMiss=True;
else: goHit=True;
elif targetHit and not targetMiss:goHit = True;
elif not targetHit and targetMiss:goMiss = True;
if goMiss:a.loc[row,"target_miss_time"] = targetMiss;
elif goHit:a.loc[row,"target_hit_time"] = targetHit;
print '#'*50
print a
'''
##################################################
price reversal target1 target2 target_hit_time target_miss_time
09:00:00 1 NaN NaN NaN NaN NaN
10:00:00 0 NaN NaN NaN NaN NaN
11:00:00 3 NaN NaN NaN NaN NaN
12:00:00 4 NaN NaN NaN NaN NaN
13:00:00 7 NaN NaN NaN NaN NaN
14:00:00 6 6.0 7.0 0.0 20:00:00 NaN
15:00:00 5 NaN NaN NaN NaN NaN
16:00:00 4 NaN NaN NaN NaN NaN
17:00:00 2 NaN NaN NaN NaN NaN
18:00:00 2 NaN NaN NaN NaN NaN
19:00:00 4 NaN NaN NaN NaN NaN
20:00:00 8 NaN NaN NaN NaN NaN
'''
这篇关于修改代码以捕获大于-而不是完全匹配的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
