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Flag only first row where condition is met in a DataFrame
Add one row to pandas DataFrameFilter dataframe rows if value in column is in a set list of valuesUse a list of values to select rows from a pandas dataframeHow to drop rows of Pandas DataFrame whose value in certain columns is NaNHow do I get the row count of a Pandas dataframe?Selecting a row of pandas series/dataframe by integer indexHow to iterate over rows in a DataFrame in Pandas?Select rows from a DataFrame based on values in a column in pandasDeleting DataFrame row in Pandas based on column valuer - reorder certain rows if condition is met
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
8
I have the following DataFrame df, which can be created as follows:
date_today = datetime.now().date()
days = pd.date_range(date_today, date_today + timedelta(19), freq='D')
x = np.arange(0,2*np.pi,0.1*np.pi) # start,stop,step
y = np.sin(x)
df = pd.DataFrame('dates': days, 'vals': y, 'is_hit': abs(y)>0.9)
df = df.set_index('dates')
And which looks like this:
is_hit vals
dates
2019-03-27 False 0.000000e+00
2019-03-28 False 3.090170e-01
2019-03-29 False 5.877853e-01
2019-03-30 False 8.090170e-01
2019-03-31 True 9.510565e-01
2019-04-01 True 1.000000e+00
2019-04-02 True 9.510565e-01
2019-04-03 False 8.090170e-01
2019-04-04 False 5.877853e-01
2019-04-05 False 3.090170e-01
2019-04-06 False 1.224647e-16
2019-04-07 False -3.090170e-01
2019-04-08 False -5.877853e-01
2019-04-09 False -8.090170e-01
2019-04-10 True -9.510565e-01
2019-04-11 True -1.000000e+00
2019-04-12 True -9.510565e-01
2019-04-13 False -8.090170e-01
2019-04-14 False -5.877853e-01
2019-04-15 False -3.090170e-01
I want to flag the rows where the is_hit condition is True for the first time, such that the expected new column hit_first would be:
is_hit vals hit_first
dates
2019-03-27 False 0.000000e+00 False
2019-03-28 False 3.090170e-01 False
2019-03-29 False 5.877853e-01 False
2019-03-30 False 8.090170e-01 False
2019-03-31 True 9.510565e-01 True
2019-04-01 True 1.000000e+00 False
2019-04-02 True 9.510565e-01 False
2019-04-03 False 8.090170e-01 False
2019-04-04 False 5.877853e-01 False
2019-04-05 False 3.090170e-01 False
2019-04-06 False 1.224647e-16 False
2019-04-07 False -3.090170e-01 False
2019-04-08 False -5.877853e-01 False
2019-04-09 False -8.090170e-01 False
2019-04-10 True -9.510565e-01 True
2019-04-11 True -1.000000e+00 False
2019-04-12 True -9.510565e-01 False
2019-04-13 False -8.090170e-01 False
2019-04-14 False -5.877853e-01 False
2019-04-15 False -3.090170e-01 False
python pandas dataframe
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
add a comment |
8
I have the following DataFrame df, which can be created as follows:
date_today = datetime.now().date()
days = pd.date_range(date_today, date_today + timedelta(19), freq='D')
x = np.arange(0,2*np.pi,0.1*np.pi) # start,stop,step
y = np.sin(x)
df = pd.DataFrame('dates': days, 'vals': y, 'is_hit': abs(y)>0.9)
df = df.set_index('dates')
And which looks like this:
is_hit vals
dates
2019-03-27 False 0.000000e+00
2019-03-28 False 3.090170e-01
2019-03-29 False 5.877853e-01
2019-03-30 False 8.090170e-01
2019-03-31 True 9.510565e-01
2019-04-01 True 1.000000e+00
2019-04-02 True 9.510565e-01
2019-04-03 False 8.090170e-01
2019-04-04 False 5.877853e-01
2019-04-05 False 3.090170e-01
2019-04-06 False 1.224647e-16
2019-04-07 False -3.090170e-01
2019-04-08 False -5.877853e-01
2019-04-09 False -8.090170e-01
2019-04-10 True -9.510565e-01
2019-04-11 True -1.000000e+00
2019-04-12 True -9.510565e-01
2019-04-13 False -8.090170e-01
2019-04-14 False -5.877853e-01
2019-04-15 False -3.090170e-01
I want to flag the rows where the is_hit condition is True for the first time, such that the expected new column hit_first would be:
is_hit vals hit_first
dates
2019-03-27 False 0.000000e+00 False
2019-03-28 False 3.090170e-01 False
2019-03-29 False 5.877853e-01 False
2019-03-30 False 8.090170e-01 False
2019-03-31 True 9.510565e-01 True
2019-04-01 True 1.000000e+00 False
2019-04-02 True 9.510565e-01 False
2019-04-03 False 8.090170e-01 False
2019-04-04 False 5.877853e-01 False
2019-04-05 False 3.090170e-01 False
2019-04-06 False 1.224647e-16 False
2019-04-07 False -3.090170e-01 False
2019-04-08 False -5.877853e-01 False
2019-04-09 False -8.090170e-01 False
2019-04-10 True -9.510565e-01 True
2019-04-11 True -1.000000e+00 False
2019-04-12 True -9.510565e-01 False
2019-04-13 False -8.090170e-01 False
2019-04-14 False -5.877853e-01 False
2019-04-15 False -3.090170e-01 False
python pandas dataframe
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
add a comment |
8
8
8
I have the following DataFrame df, which can be created as follows:
date_today = datetime.now().date()
days = pd.date_range(date_today, date_today + timedelta(19), freq='D')
x = np.arange(0,2*np.pi,0.1*np.pi) # start,stop,step
y = np.sin(x)
df = pd.DataFrame('dates': days, 'vals': y, 'is_hit': abs(y)>0.9)
df = df.set_index('dates')
And which looks like this:
is_hit vals
dates
2019-03-27 False 0.000000e+00
2019-03-28 False 3.090170e-01
2019-03-29 False 5.877853e-01
2019-03-30 False 8.090170e-01
2019-03-31 True 9.510565e-01
2019-04-01 True 1.000000e+00
2019-04-02 True 9.510565e-01
2019-04-03 False 8.090170e-01
2019-04-04 False 5.877853e-01
2019-04-05 False 3.090170e-01
2019-04-06 False 1.224647e-16
2019-04-07 False -3.090170e-01
2019-04-08 False -5.877853e-01
2019-04-09 False -8.090170e-01
2019-04-10 True -9.510565e-01
2019-04-11 True -1.000000e+00
2019-04-12 True -9.510565e-01
2019-04-13 False -8.090170e-01
2019-04-14 False -5.877853e-01
2019-04-15 False -3.090170e-01
I want to flag the rows where the is_hit condition is True for the first time, such that the expected new column hit_first would be:
is_hit vals hit_first
dates
2019-03-27 False 0.000000e+00 False
2019-03-28 False 3.090170e-01 False
2019-03-29 False 5.877853e-01 False
2019-03-30 False 8.090170e-01 False
2019-03-31 True 9.510565e-01 True
2019-04-01 True 1.000000e+00 False
2019-04-02 True 9.510565e-01 False
2019-04-03 False 8.090170e-01 False
2019-04-04 False 5.877853e-01 False
2019-04-05 False 3.090170e-01 False
2019-04-06 False 1.224647e-16 False
2019-04-07 False -3.090170e-01 False
2019-04-08 False -5.877853e-01 False
2019-04-09 False -8.090170e-01 False
2019-04-10 True -9.510565e-01 True
2019-04-11 True -1.000000e+00 False
2019-04-12 True -9.510565e-01 False
2019-04-13 False -8.090170e-01 False
2019-04-14 False -5.877853e-01 False
2019-04-15 False -3.090170e-01 False
python pandas dataframe
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
I have the following DataFrame df, which can be created as follows:
date_today = datetime.now().date()
days = pd.date_range(date_today, date_today + timedelta(19), freq='D')
x = np.arange(0,2*np.pi,0.1*np.pi) # start,stop,step
y = np.sin(x)
df = pd.DataFrame('dates': days, 'vals': y, 'is_hit': abs(y)>0.9)
df = df.set_index('dates')
And which looks like this:
is_hit vals
dates
2019-03-27 False 0.000000e+00
2019-03-28 False 3.090170e-01
2019-03-29 False 5.877853e-01
2019-03-30 False 8.090170e-01
2019-03-31 True 9.510565e-01
2019-04-01 True 1.000000e+00
2019-04-02 True 9.510565e-01
2019-04-03 False 8.090170e-01
2019-04-04 False 5.877853e-01
2019-04-05 False 3.090170e-01
2019-04-06 False 1.224647e-16
2019-04-07 False -3.090170e-01
2019-04-08 False -5.877853e-01
2019-04-09 False -8.090170e-01
2019-04-10 True -9.510565e-01
2019-04-11 True -1.000000e+00
2019-04-12 True -9.510565e-01
2019-04-13 False -8.090170e-01
2019-04-14 False -5.877853e-01
2019-04-15 False -3.090170e-01
I want to flag the rows where the is_hit condition is True for the first time, such that the expected new column hit_first would be:
is_hit vals hit_first
dates
2019-03-27 False 0.000000e+00 False
2019-03-28 False 3.090170e-01 False
2019-03-29 False 5.877853e-01 False
2019-03-30 False 8.090170e-01 False
2019-03-31 True 9.510565e-01 True
2019-04-01 True 1.000000e+00 False
2019-04-02 True 9.510565e-01 False
2019-04-03 False 8.090170e-01 False
2019-04-04 False 5.877853e-01 False
2019-04-05 False 3.090170e-01 False
2019-04-06 False 1.224647e-16 False
2019-04-07 False -3.090170e-01 False
2019-04-08 False -5.877853e-01 False
2019-04-09 False -8.090170e-01 False
2019-04-10 True -9.510565e-01 True
2019-04-11 True -1.000000e+00 False
2019-04-12 True -9.510565e-01 False
2019-04-13 False -8.090170e-01 False
2019-04-14 False -5.877853e-01 False
2019-04-15 False -3.090170e-01 False
python pandas dataframe
python pandas dataframe
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
asked Mar 27 at 12:23
JejeBelfortJejeBelfort
6911624
6911624
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered Mar 27 at 12:28
ecortazarecortazar
96618
add a comment |
3
Use Series.shift chained with & for bitwise AND:
df['hit_first'] = df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
print (df)
vals is_hit hit_first
dates
2019-03-27 0.000000e+00 False False
2019-03-28 3.090170e-01 False False
2019-03-29 5.877853e-01 False False
2019-03-30 8.090170e-01 False False
2019-03-31 9.510565e-01 True True
2019-04-01 1.000000e+00 True False
2019-04-02 9.510565e-01 True False
2019-04-03 8.090170e-01 False False
2019-04-04 5.877853e-01 False False
2019-04-05 3.090170e-01 False False
2019-04-06 1.224647e-16 False False
2019-04-07 -3.090170e-01 False False
2019-04-08 -5.877853e-01 False False
2019-04-09 -8.090170e-01 False False
2019-04-10 -9.510565e-01 True True
2019-04-11 -1.000000e+00 True False
2019-04-12 -9.510565e-01 True False
2019-04-13 -8.090170e-01 False False
2019-04-14 -5.877853e-01 False False
2019-04-15 -3.090170e-01 False False
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
add a comment |
3
I also, think you can do it this way:
df['is_hit'].astype(int).diff() == 1
Output:
dates
2019-03-27 False
2019-03-28 False
2019-03-29 False
2019-03-30 False
2019-03-31 True
2019-04-01 False
2019-04-02 False
2019-04-03 False
2019-04-04 False
2019-04-05 False
2019-04-06 False
2019-04-07 False
2019-04-08 False
2019-04-09 False
2019-04-10 True
2019-04-11 False
2019-04-12 False
2019-04-13 False
2019-04-14 False
2019-04-15 False
Name: is_hit, dtype: bool
Timings:
%timeit df['is_hit'] & (~df['is_hit']).shift(1)
1.13 ms ± 5.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
908 µs ± 9.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].astype(int).diff() == 1
689 µs ± 8.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
add a comment |
-1
Also this can be done by using simple difference between the series and it's shifted series by 1 period :
df['hit_first'] = df['is_hit']-df['is_hit'].shift()==1
answered Mar 27 at 12:54
LoochieLoochie
959310
1
The use ofnp.wherehere is quite pointless.
– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered Mar 27 at 12:28
ecortazarecortazar
96618
add a comment |
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered Mar 27 at 12:28
ecortazarecortazar
96618
add a comment |
10
10
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered Mar 27 at 12:28
ecortazarecortazar
96618
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered Mar 27 at 12:28
ecortazarecortazar
96618
answered Mar 27 at 12:28
ecortazarecortazar
96618
answered Mar 27 at 12:28
ecortazarecortazar
96618
answered Mar 27 at 12:28
ecortazarecortazar
96618
96618
add a comment |
add a comment |
3
Use Series.shift chained with & for bitwise AND:
df['hit_first'] = df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
print (df)
vals is_hit hit_first
dates
2019-03-27 0.000000e+00 False False
2019-03-28 3.090170e-01 False False
2019-03-29 5.877853e-01 False False
2019-03-30 8.090170e-01 False False
2019-03-31 9.510565e-01 True True
2019-04-01 1.000000e+00 True False
2019-04-02 9.510565e-01 True False
2019-04-03 8.090170e-01 False False
2019-04-04 5.877853e-01 False False
2019-04-05 3.090170e-01 False False
2019-04-06 1.224647e-16 False False
2019-04-07 -3.090170e-01 False False
2019-04-08 -5.877853e-01 False False
2019-04-09 -8.090170e-01 False False
2019-04-10 -9.510565e-01 True True
2019-04-11 -1.000000e+00 True False
2019-04-12 -9.510565e-01 True False
2019-04-13 -8.090170e-01 False False
2019-04-14 -5.877853e-01 False False
2019-04-15 -3.090170e-01 False False
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
add a comment |
3
Use Series.shift chained with & for bitwise AND:
df['hit_first'] = df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
print (df)
vals is_hit hit_first
dates
2019-03-27 0.000000e+00 False False
2019-03-28 3.090170e-01 False False
2019-03-29 5.877853e-01 False False
2019-03-30 8.090170e-01 False False
2019-03-31 9.510565e-01 True True
2019-04-01 1.000000e+00 True False
2019-04-02 9.510565e-01 True False
2019-04-03 8.090170e-01 False False
2019-04-04 5.877853e-01 False False
2019-04-05 3.090170e-01 False False
2019-04-06 1.224647e-16 False False
2019-04-07 -3.090170e-01 False False
2019-04-08 -5.877853e-01 False False
2019-04-09 -8.090170e-01 False False
2019-04-10 -9.510565e-01 True True
2019-04-11 -1.000000e+00 True False
2019-04-12 -9.510565e-01 True False
2019-04-13 -8.090170e-01 False False
2019-04-14 -5.877853e-01 False False
2019-04-15 -3.090170e-01 False False
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
add a comment |
3
3
3
Use Series.shift chained with & for bitwise AND:
df['hit_first'] = df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
print (df)
vals is_hit hit_first
dates
2019-03-27 0.000000e+00 False False
2019-03-28 3.090170e-01 False False
2019-03-29 5.877853e-01 False False
2019-03-30 8.090170e-01 False False
2019-03-31 9.510565e-01 True True
2019-04-01 1.000000e+00 True False
2019-04-02 9.510565e-01 True False
2019-04-03 8.090170e-01 False False
2019-04-04 5.877853e-01 False False
2019-04-05 3.090170e-01 False False
2019-04-06 1.224647e-16 False False
2019-04-07 -3.090170e-01 False False
2019-04-08 -5.877853e-01 False False
2019-04-09 -8.090170e-01 False False
2019-04-10 -9.510565e-01 True True
2019-04-11 -1.000000e+00 True False
2019-04-12 -9.510565e-01 True False
2019-04-13 -8.090170e-01 False False
2019-04-14 -5.877853e-01 False False
2019-04-15 -3.090170e-01 False False
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
Use Series.shift chained with & for bitwise AND:
df['hit_first'] = df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
print (df)
vals is_hit hit_first
dates
2019-03-27 0.000000e+00 False False
2019-03-28 3.090170e-01 False False
2019-03-29 5.877853e-01 False False
2019-03-30 8.090170e-01 False False
2019-03-31 9.510565e-01 True True
2019-04-01 1.000000e+00 True False
2019-04-02 9.510565e-01 True False
2019-04-03 8.090170e-01 False False
2019-04-04 5.877853e-01 False False
2019-04-05 3.090170e-01 False False
2019-04-06 1.224647e-16 False False
2019-04-07 -3.090170e-01 False False
2019-04-08 -5.877853e-01 False False
2019-04-09 -8.090170e-01 False False
2019-04-10 -9.510565e-01 True True
2019-04-11 -1.000000e+00 True False
2019-04-12 -9.510565e-01 True False
2019-04-13 -8.090170e-01 False False
2019-04-14 -5.877853e-01 False False
2019-04-15 -3.090170e-01 False False
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
edited Mar 27 at 12:33
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
answered Mar 27 at 12:28
jezraeljezrael
356k26320396
356k26320396
add a comment |
add a comment |
3
I also, think you can do it this way:
df['is_hit'].astype(int).diff() == 1
Output:
dates
2019-03-27 False
2019-03-28 False
2019-03-29 False
2019-03-30 False
2019-03-31 True
2019-04-01 False
2019-04-02 False
2019-04-03 False
2019-04-04 False
2019-04-05 False
2019-04-06 False
2019-04-07 False
2019-04-08 False
2019-04-09 False
2019-04-10 True
2019-04-11 False
2019-04-12 False
2019-04-13 False
2019-04-14 False
2019-04-15 False
Name: is_hit, dtype: bool
Timings:
%timeit df['is_hit'] & (~df['is_hit']).shift(1)
1.13 ms ± 5.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
908 µs ± 9.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].astype(int).diff() == 1
689 µs ± 8.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
add a comment |
3
I also, think you can do it this way:
df['is_hit'].astype(int).diff() == 1
Output:
dates
2019-03-27 False
2019-03-28 False
2019-03-29 False
2019-03-30 False
2019-03-31 True
2019-04-01 False
2019-04-02 False
2019-04-03 False
2019-04-04 False
2019-04-05 False
2019-04-06 False
2019-04-07 False
2019-04-08 False
2019-04-09 False
2019-04-10 True
2019-04-11 False
2019-04-12 False
2019-04-13 False
2019-04-14 False
2019-04-15 False
Name: is_hit, dtype: bool
Timings:
%timeit df['is_hit'] & (~df['is_hit']).shift(1)
1.13 ms ± 5.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
908 µs ± 9.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].astype(int).diff() == 1
689 µs ± 8.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
add a comment |
3
3
3
I also, think you can do it this way:
df['is_hit'].astype(int).diff() == 1
Output:
dates
2019-03-27 False
2019-03-28 False
2019-03-29 False
2019-03-30 False
2019-03-31 True
2019-04-01 False
2019-04-02 False
2019-04-03 False
2019-04-04 False
2019-04-05 False
2019-04-06 False
2019-04-07 False
2019-04-08 False
2019-04-09 False
2019-04-10 True
2019-04-11 False
2019-04-12 False
2019-04-13 False
2019-04-14 False
2019-04-15 False
Name: is_hit, dtype: bool
Timings:
%timeit df['is_hit'] & (~df['is_hit']).shift(1)
1.13 ms ± 5.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
908 µs ± 9.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].astype(int).diff() == 1
689 µs ± 8.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
I also, think you can do it this way:
df['is_hit'].astype(int).diff() == 1
Output:
dates
2019-03-27 False
2019-03-28 False
2019-03-29 False
2019-03-30 False
2019-03-31 True
2019-04-01 False
2019-04-02 False
2019-04-03 False
2019-04-04 False
2019-04-05 False
2019-04-06 False
2019-04-07 False
2019-04-08 False
2019-04-09 False
2019-04-10 True
2019-04-11 False
2019-04-12 False
2019-04-13 False
2019-04-14 False
2019-04-15 False
Name: is_hit, dtype: bool
Timings:
%timeit df['is_hit'] & (~df['is_hit']).shift(1)
1.13 ms ± 5.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].ne(df['is_hit'].shift()) & df['is_hit']
908 µs ± 9.53 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit df['is_hit'].astype(int).diff() == 1
689 µs ± 8.24 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
edited Mar 27 at 12:57
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
answered Mar 27 at 12:50
Scott BostonScott Boston
58k73258
58k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
add a comment |
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
2
2
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
Nice, maybe performance in large data should be interesting.
– jezrael
Mar 27 at 13:11
add a comment |
-1
Also this can be done by using simple difference between the series and it's shifted series by 1 period :
df['hit_first'] = df['is_hit']-df['is_hit'].shift()==1
answered Mar 27 at 12:54
LoochieLoochie
959310
1
The use ofnp.wherehere is quite pointless.
– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
add a comment |
-1
Also this can be done by using simple difference between the series and it's shifted series by 1 period :
df['hit_first'] = df['is_hit']-df['is_hit'].shift()==1
answered Mar 27 at 12:54
LoochieLoochie
959310
1
The use ofnp.wherehere is quite pointless.
– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
add a comment |
-1
-1
-1
Also this can be done by using simple difference between the series and it's shifted series by 1 period :
df['hit_first'] = df['is_hit']-df['is_hit'].shift()==1
answered Mar 27 at 12:54
LoochieLoochie
959310
Also this can be done by using simple difference between the series and it's shifted series by 1 period :
df['hit_first'] = df['is_hit']-df['is_hit'].shift()==1
answered Mar 27 at 12:54
LoochieLoochie
959310
edited Mar 27 at 22:45
answered Mar 27 at 12:54
LoochieLoochie
959310
answered Mar 27 at 12:54
LoochieLoochie
959310
answered Mar 27 at 12:54
LoochieLoochie
959310
959310
1
The use ofnp.wherehere is quite pointless.
– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
add a comment |
1
The use ofnp.wherehere is quite pointless.
– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
1
1
The use of
np.where here is quite pointless.– miradulo
Mar 27 at 18:50
The use of
np.where here is quite pointless.– miradulo
Mar 27 at 18:50
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
Yes I understood. Thanks :)
– Loochie
Mar 27 at 20:42
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– DebanjanB
Mar 27 at 21:22
add a comment |
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-dataframe, pandas, python
