Let's label a dataframe with two columns, A,B, and 100M rows. Starting at the index i
, we want to know if the data in column B is trending down or trending up comparing to the data at [i, 'A']
.
Here is a loop:
import pandas as pd
df = pd.DataFrame({'A': [0,1,2,3,5,0,0,0,0,0], 'B': [1, 10, -10, 2, 3,0,0,0,0,0], "label":[0,0,0,0,0,0,0,0,0,0]})
for i in range (0,5):
j = i
while j in range (i,i 5) and df.at[i,'label'] == 0: #if classfied, no need to continue
if df.at[j,'B']-df.at[i,'A']>= 10:
df.at[i,'label'] = 1 #Label 1 means trending up
if df.at[j,'B']-df.at[i,'A']<= -10:
df.at[i,'label'] = 2 #Label 2 means trending down
j=j 1
[out]
A B label
0 1 1
1 10 2
2 -10 2
3 2 0
5 3 0
...
The estimated finishing time for this code is 30 days. (A human with a plot and a ruler might finish this task faster.)
What is a fast way to do this? Ideally without a loop.
CodePudding user response:
Looping on Dataframe is slow compared to using Pandas methods.
The task can be accomplished using Pandas vectorized methods:
- rolling method which does computations in a rolling window
- min & max methods which we compute in the rolling window
- where method DataFrame where allows us to set values based upon logic
Code
def set_trend(df, threshold = 10, window_size = 2):
'''
Use rolling_window to find max/min values in a window from the current point
rolling window normally looks at backward values
We use technique from https://stackoverflow.com/questions/22820292/how-to-use-pandas-rolling-functions-on-a-forward-looking-basis/22820689#22820689
to look at forward values
'''
# To have a rolling window on lookahead values in column B
# We reverse values in column B
df['B_rev'] = df["B"].values[::-1]
# Max & Min in B_rev, then reverse order of these max/min
# https://stackoverflow.com/questions/50837012/pandas-rolling-min-max
df['max_'] = df.B_rev.rolling(window_size, min_periods = 0).max().values[::-1]
df['min_'] = df.B_rev.rolling(window_size, min_periods = 0).min().values[::-1]
nrows = df.shape[0] - 1 # adjustment for argmax & armin indexes since rows are in reverse order
# i.e. idx = nrows - x.argmax() give index for max in non-reverse row
df['max_idx'] = df.B_rev.rolling(window_size, min_periods = 0).apply(lambda x: nrows - x.argmax(), raw = True).values[::-1]
df['min_idx'] = df.B_rev.rolling(window_size, min_periods = 0).apply(lambda x: nrows - x.argmin(), raw = True).values[::-1]
# Use np.select to implement label assignment logic
conditions = [
(df['max_'] - df["A"] >= threshold) & (df['max_idx'] <= df['min_idx']), # max above & comes first
(df['min_'] - df["A"] <= -threshold) & (df['min_idx'] <= df['max_idx']), # min below & comes first
df['max_'] - df["A"] >= threshold, # max above threshold but didn't come first
df['min_'] - df["A"] <= -threshold, # min below threshold but didn't come first
]
choices = [
1, # max above & came first
2, # min above & came first
1, # max above threshold
2, # min above threshold
]
df['label'] = np.select(conditions, choices, default = 0)
# Drop scratch computation columns
df.drop(['B_rev', 'max_', 'min_', 'max_idx', 'min_idx'], axis = 1, inplace = True)
return df
Tests
Case 1
df = pd.DataFrame({'A': [0,1,2,3,5,0,0,0,0,0], 'B': [1, 10, -10, 2, 3,0,0,0,0,0], "label":[0,0,0,0,0,0,0,0,0,0]})
display(set_trend(df, 10, 4))
Case 2
df = pd.DataFrame({'A': [0,1,2], 'B': [1, -10, 10]})
display(set_trend(df, 10, 4))
Output
Case 1
A B label
0 0 1 1
1 1 10 2
2 2 -10 2
3 3 2 0
4 5 3 0
5 0 0 0
6 0 0 0
7 0 0 0
8 0 0 0
9 0 0 0
Case 2
A B label
0 0 1 2
1 1 -10 2
2 2 10 0