I have the following dataframe:
True_False cum_val
Date
2018-01-02 False NaN
2018-01-03 False 0.006399
2018-01-04 False 0.010427
2018-01-05 False 0.017461
2018-01-08 False 0.019124
2018-01-09 False 0.020426
2018-01-10 False 0.019314
2018-01-11 False 0.026348
2018-01-12 False 0.033098
2018-01-16 False 0.029573
2018-01-17 False 0.038988
2018-01-18 False 0.037372
2018-01-19 False 0.041757
2018-01-22 False 0.049824
2018-01-23 False 0.051998
2018-01-24 False 0.051438
2018-01-25 False 0.052041
2018-01-26 False 0.063882
2018-01-29 False 0.057150
2018-01-30 True -0.010899
2018-01-31 True -0.010410
2018-02-01 True -0.011058
2018-02-02 True -0.032266
2018-02-05 True -0.073246
2018-02-06 True -0.055805
2018-02-07 True -0.060806
2018-02-08 True -0.098343
2018-02-09 True -0.083407
2018-02-12 False 0.013915
2018-02-13 False 0.016528
2018-02-14 False 0.029930
2018-02-15 False 0.041999
2018-02-16 False 0.042373
2018-02-20 False 0.036531
2018-02-21 False 0.031035
2018-03-06 False 0.013671
How can I drop the row second value after False
all the the True
values till the second True Value
till the second False
?
Such as for example:
True_False cum_val
Date
2020-01-21 False 0.022808
2020-01-22 False 0.023097
2020-01-23 True 0.001141
2020-01-24 True -0.007901 # <- Start drop here since this is the second True
2020-01-27 True -0.023632
2020-01-28 False -0.013578
2020-01-29 False -0.000867 #< - End Drop Here Since this is the second False
2020-01-30 False 0.003134
Edit 1:
I would like to add 1 more condition on the new df:
2020-01-22 0.000289 False
2020-01-23 0.001141 True
2020-01-27 -0.015731 True # <- Start Drop Here
2020-01-28 0.010054 True
2020-01-29 -0.000867 False
2020-01-30 0.003134 True #<-End drop here
2020-02-03 0.007255 True
As you have mentioned in the comment: [True, True, True, False, True]
In this case it would still start the drop at the second True
value but would stop the drop right after the first False
even though the second value has toggled to True
. If the next value is still True
drop it till the value after False
We can use the following syntax to drop rows in a pandas DataFrame based on condition: Note: We can also use the drop () function to drop rows from a DataFrame, but this function has been shown to be much slower than just assigning the DataFrame to a filtered version of itself.
Drop Rows with Duplicate in pandas. Delete or Drop rows with condition in python pandas using drop () function. Drop rows by index / position in pandas. Drop NA rows or missing rows in pandas python. labels: String or list of strings referring row. axis: int or string value, 0 ‘index’ for Rows and 1 ‘columns’ for Columns.
Alternatively, you can also try another most used approach to drop rows by condition using loc [] and df []. Note that these methods actually filter the data, by negating this you will get the desired output. # Remove row df2 = df [ df. Fee >= 24000] print( df2) #Using loc [] df2 = df. loc [ df ["Fee"] >= 24000 ] print( df2)
Sometimes you have to remove rows from dataframe based on some specific condition. It can be done by passing the condition df [your_conditon] inside the drop () method. For example, I want to drop rows that have a value greater than 4 of Column A.
Let's try using where
with ffill
and parameter limit=2
then boolean filtering:
df[~(df['True_False'].where(df['True_False']).ffill(limit=2).cumsum() > 1)]
Output:
| | Date | True_False | cum_val |
|----|------------|--------------|-----------|
| 0 | 2020-01-21 | False | 1 |
| 1 | 2020-01-22 | False | 2 |
| 2 | 2020-01-23 | True | 3 |
| 7 | 2020-01-28 | False | 8 |
Details:
where
ffill(limit=2)
cumsum
so we can add consecutive True and select
those greater than 2 Here's what I tried. The data I created is:
Date True_False cum_val
0 2020-01-21 False 1
1 2020-01-22 False 2
2 2020-01-23 True 3
3 2020-01-24 True 4
4 2020-01-25 True 5
5 2020-01-26 False 6
6 2020-01-27 False 7
7 2020-01-28 False 8
true_count = 0
false_count = 0
drop_continue = False
for index, row in df.iterrows():
if row['True_False'] is True and drop_continue is False:
true_count +=1
if true_count == 2:
drop_continue = True
df.drop(index, inplace=True)
true_count = 0
continue
if drop_continue is True:
if row['True_False'] is True:
df.drop(index, inplace=True)
if row['True_False'] is False:
false_count += 1
if false_count <2:
df.drop(index, inplace=True)
else:
drop_continue = False
false_count = 0
Output
Date True_False cum_val
0 2020-01-21 False 1
1 2020-01-22 False 2
2 2020-01-23 True 3
6 2020-01-27 False 7
7 2020-01-28 False 8
You could use Series.Shift
and Series.bfill
:
df = df[~df['True_False'].shift().bfill()]
print(df)
Date True_False cum_val
0 2020-01-21 False 0.022808
1 2020-01-22 False 0.023097
2 2020-01-23 True 0.001141
6 2020-01-29 False -0.000867
7 2020-01-30 False 0.003134
You can do:
#mark start of the area you want to drop
df["dropit"]=np.where(df["True_False"] & df["True_False"].shift(1) & np.logical_not(df["True_False"].shift(2)), "start", None)
#mark the end of the drop area
df["dropit"]=np.where(np.logical_not(df["True_False"].shift(1)) & df["True_False"].shift(2), "end", df["dropit"])
#indicate gaps between the different drop areas:
df.loc[df["dropit"].shift().eq("end")&df["dropit"].ne("start"), "dropit"]="keep"
#forward fill
df["dropit"]=df["dropit"].ffill()
#drop marked drop areas and drop "dropit" column
df=df.drop(df.loc[df["dropit"].isin(["start", "end"])].index, axis=0).drop("dropit", axis=1)
Outputs:
True_False cum_val
Date
2018-01-02 False NaN
2018-01-03 False 0.006399
2018-01-04 False 0.010427
2018-01-05 False 0.017461
2018-01-08 False 0.019124
2018-01-09 False 0.020426
2018-01-10 False 0.019314
2018-01-11 False 0.026348
2018-01-12 False 0.033098
2018-01-16 False 0.029573
2018-01-17 False 0.038988
2018-01-18 False 0.037372
2018-01-19 False 0.041757
2018-01-22 False 0.049824
2018-01-23 False 0.051998
2018-01-24 False 0.051438
2018-01-25 False 0.052041
2018-01-26 False 0.063882
2018-01-29 False 0.057150
2018-01-30 True -0.010899
2018-02-14 False 0.029930
2018-02-15 False 0.041999
2018-02-16 False 0.042373
2018-02-20 False 0.036531
2018-02-21 False 0.031035
2018-03-06 False 0.013671
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