Complex Excel Formula in Pandas - pandas

Question

Excel Formulas I am trying to replicate in pandas:
Click here to download workbook
* Look at columns D, E and F
entsig and exsig are manual and can be changed. In real life they would be derived from the value of another column or a comparison of two other columns
ent = 1 if entsig previous = 1 and in = 0
in = 1 if ent previous = 1 or (in previous = 1 and ex = 0)
ex = 1 if exsig previous = 1 and in previous = 1
so either ent, in, or ex will always be = 1 but never more than one of them
import pandas as pd
df = pd.DataFrame(
[[0,0,0,0,0], [1,0,0,0,0], [1,0,0,0,0], [1,0,0,0,0], [0,0,0,0,0],
[0,1,0,0,0], [0,1,0,0,0], [1,0,0,0,0], [1,0,0,0,0], [0,0,0,0,0],
[0,0,0,0,0], [0,0,0,0,0], [0,1,0,0,0], [0,1,0,0,0], [0,1,0,0,0],
[0,0,0,0,0], [0,0,0,0,0], [1,0,0,0,0], [1,0,0,0,0], [1,0,0,0,0],
[1,1,0,0,0], [0,1,0,0,0], [0,1,0,0,0], [0,1,0,0,0]],
columns=['entsig', 'exsig','ent', 'in', 'ex'])
for i in df.index:
df['ent'][(df.entsig.shift(1)==1) & (df['ent'].shift(1) == 0) & (df['in'].shift(1) == 0)]=1
df['ex'][(df.exsig.shift(1)==1) & (df['in'].shift(1)==1)]=1
df['in'][(df.ent.shift(1)==1) | ((df['in'].shift(1)==1) & (df['ex']==0))]=1
for j in df.index:
df['ent'][df['in'] == 1]=0
df['in'][df['ex']==1]=0
df['ex'][df['ex'].shift(1)==1]=0
df
results in
entsig exsig ent in ex
0 0 0 0 0 0
1 1 0 0 0 0
2 1 0 1 0 0
3 1 0 0 1 0
4 0 0 0 1 0
5 0 1 0 1 0
6 0 1 0 0 1
7 1 0 0 0 0
8 1 0 1 0 0
9 0 0 0 1 0
10 0 0 0 1 0
11 0 0 0 1 0
12 0 1 0 1 0
13 0 1 0 0 1
14 0 1 0 0 0
15 0 0 0 0 0
16 0 0 0 0 0
17 1 0 0 0 0
18 1 0 1 0 0
19 1 0 0 1 0
20 1 1 0 1 0
21 0 1 0 0 1
22 0 1 0 0 0
23 0 1 0 0 0
Question
How can I make this code faster? It runs slow because it's a loop but I have not been able to come up with a solution that does not use loops. Any ideas or comments are appreciated.

Answer 1

If we can assume every group of 1's in entsig is followed by at least one 1 in
exsig, then you could compute ent, ex and in like this:
def ent_in_ex(df):
entsig_mask = (df['entsig'].diff().shift(1) == 1)
exsig_mask = (df['exsig'].diff().shift(1) == 1)
df.loc[entsig_mask, 'ent'] = 1
df.loc[exsig_mask, 'ex'] = 1
df['in'] = df['ent'].shift(1).cumsum().subtract(df['ex'].cumsum(), fill_value=0)
return df
If we can make this assumption, then ent_in_ex is significantly faster:
In [5]: %timeit orig(df)
10 loops, best of 3: 185 ms per loop
In [6]: %timeit ent_in_ex(df)
100 loops, best of 3: 2.23 ms per loop
In [95]: orig(df).equals(ent_in_ex(df))
Out[95]: True
where orig is the original code:
def orig(df):
for i in df.index:
df['ent'][(df.entsig.shift(1)==1) & (df['ent'].shift(1) == 0) & (df['in'].shift(1) == 0)]=1
df['ex'][(df.exsig.shift(1)==1) & (df['in'].shift(1)==1)]=1
df['in'][(df.ent.shift(1)==1) | ((df['in'].shift(1)==1) & (df['ex']==0))]=1
for j in df.index:
df['ent'][df['in'] == 1]=0
df['in'][df['ex']==1]=0
df['ex'][df['ex'].shift(1)==1]=0
return df