Flag only first row where condition is met in a DataFrameAdd 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
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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
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 18 hours ago
JejeBelfortJejeBelfort
6591624
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 18 hours ago
JejeBelfortJejeBelfort
6591624
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 18 hours ago
JejeBelfortJejeBelfort
6591624
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 18 hours ago
JejeBelfortJejeBelfort
6591624
asked 18 hours ago
JejeBelfortJejeBelfort
6591624
asked 18 hours ago
JejeBelfortJejeBelfort
6591624
asked 18 hours ago
JejeBelfortJejeBelfort
6591624
asked 18 hours ago
JejeBelfortJejeBelfort
6591624
6591624
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 17 hours ago
ecortazarecortazar
7867
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 17 hours ago
jezraeljezrael
352k26316391
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 17 hours ago
Scott BostonScott Boston
57.1k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
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 17 hours ago
LoochieLoochie
776310
1
The use ofnp.wherehere is quite pointless.
– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
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 17 hours ago
ecortazarecortazar
7867
add a comment |
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered 17 hours ago
ecortazarecortazar
7867
add a comment |
10
10
10
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered 17 hours ago
ecortazarecortazar
7867
My suggestion:
df['hit_first'] = df['is_hit'] & (~df['is_hit']).shift(1)
answered 17 hours ago
ecortazarecortazar
7867
answered 17 hours ago
ecortazarecortazar
7867
answered 17 hours ago
ecortazarecortazar
7867
answered 17 hours ago
ecortazarecortazar
7867
7867
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 17 hours ago
jezraeljezrael
352k26316391
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 17 hours ago
jezraeljezrael
352k26316391
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 17 hours ago
jezraeljezrael
352k26316391
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 17 hours ago
jezraeljezrael
352k26316391
edited 17 hours ago
answered 17 hours ago
jezraeljezrael
352k26316391
answered 17 hours ago
jezraeljezrael
352k26316391
answered 17 hours ago
jezraeljezrael
352k26316391
352k26316391
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 17 hours ago
Scott BostonScott Boston
57.1k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
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 17 hours ago
Scott BostonScott Boston
57.1k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
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 17 hours ago
Scott BostonScott Boston
57.1k73258
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 17 hours ago
Scott BostonScott Boston
57.1k73258
edited 17 hours ago
answered 17 hours ago
Scott BostonScott Boston
57.1k73258
answered 17 hours ago
Scott BostonScott Boston
57.1k73258
answered 17 hours ago
Scott BostonScott Boston
57.1k73258
57.1k73258
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
add a comment |
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
2
2
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
Nice, maybe performance in large data should be interesting.
– jezrael
17 hours ago
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 17 hours ago
LoochieLoochie
776310
1
The use ofnp.wherehere is quite pointless.
– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
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 17 hours ago
LoochieLoochie
776310
1
The use ofnp.wherehere is quite pointless.
– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
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 17 hours ago
LoochieLoochie
776310
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 17 hours ago
LoochieLoochie
776310
edited 7 hours ago
answered 17 hours ago
LoochieLoochie
776310
answered 17 hours ago
LoochieLoochie
776310
answered 17 hours ago
LoochieLoochie
776310
776310
1
The use ofnp.wherehere is quite pointless.
– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
add a comment |
1
The use ofnp.wherehere is quite pointless.
– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
1
1
The use of
np.where here is quite pointless.– miradulo
11 hours ago
The use of
np.where here is quite pointless.– miradulo
11 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
Yes I understood. Thanks :)
– Loochie
9 hours ago
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
9 hours ago
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
9 hours ago
add a comment |
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