Steven G Steven G - 1 month ago 6
Python Question

Pythonic way of creating a recursive series of 1 and 0 based on a pandas time series level

I am trying to clean up a code where I have a dataframe such has:

df = pd.DataFrame({'value': {'2016-09-21': 13.30,
'2016-09-22': 12.02,
'2016-09-23': 12.28,
'2016-09-26': 14.5,
'2016-09-27': 13.1,
'2016-09-28': 12.39,
'2016-09-29': 14.02}})


I have a ON and OFF signal based on levels. when 'value' cross upward 14.39 I want to have 1 until it cross 12.50 downward such has:

df
value sig
2016-09-21 13.3000 0
2016-09-22 12.0200 0
2016-09-23 12.2800 0
2016-09-26 14.5000 1
2016-09-27 13.1000 1
2016-09-28 12.3900 0
2016-09-29 14.0200 0


I am approching the problem through a loop but I am pretty sure there's a better way to do it. here is my approach:

off, on, sig = 14.39, 12.50, 0
log = []
for level in df.itertuples():
if level.value > off:
sig = 1
elif (sig == 1) & (level.value < on):
sig = 0
log.append([level.value, sig])
log = pd.DataFrame(log, index=df.index, columns=['value', 'sig'])

Answer

Here is a vectorized solution with pandas.Series.where method:

import numpy as np

ON, OFF = 14.39, 12.50
df['sig'] = 0                                 #  set the initial value to be 0
df['sig'] = (df.sig.where(df.value < ON, 1)   #  if value > ON, set it 1
                   .where((df.value < OFF) | (df.value > ON), np.nan)  
                                              #  if value < ON, and value > OFF, set it nan
                   .ffill().fillna(0))        # forward fill the nan value as they depend 
                                              # on their previous state, and fill initial 
                                              # value as 0
df

#           value   sig
#2016-09-21 13.30     0
#2016-09-22 12.02     0
#2016-09-23 12.28     0
#2016-09-26 14.50     1
#2016-09-27 13.10     1
#2016-09-28 12.39     0
#2016-09-29 14.02     0

A similar np.where() method with maybe clearer intention´╝Ü

import numpy as np
df['sig'] = np.where(df.value > ON, 1, np.where(df.value < OFF, 0, np.nan))
df['sig'] = df.sig.ffill().fillna(0) 
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