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How to reduce program execution time by replacing for loop in pandas
How do I get the row count of a pandas DataFrame?How to iterate over rows in a DataFrame in Pandas?How to deal with SettingWithCopyWarning in Pandas?Loop through csv with Pandas, specific columnCan't execute Python Pandas set_valueDifficulty replacing values in Pandas columnAssitance needed in python pandas to reduce lines of code and cycle timeHow to index a pandas data frame starting at n?How do I sort a pandas data frame by date timeHow to loop through Date, Time and values in Pandas
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
3
I have 12000 csv files every file have 6000 rows . i am using for loop in my code , i think because of this my code execution time increased. if anyone know how to change this piece of code in to pandas package that reduce execution time
for i in range(len(df)):
if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('EOG')
df['EOG_flag'].values[i]=1
if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('gust')
df['Gust_flag'].values[i]=1
Note: this code is working well , just execution time is high
python-3.x pandas
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
add a comment |
3
I have 12000 csv files every file have 6000 rows . i am using for loop in my code , i think because of this my code execution time increased. if anyone know how to change this piece of code in to pandas package that reduce execution time
for i in range(len(df)):
if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('EOG')
df['EOG_flag'].values[i]=1
if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('gust')
df['Gust_flag'].values[i]=1
Note: this code is working well , just execution time is high
python-3.x pandas
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
add a comment |
3
3
3
0
I have 12000 csv files every file have 6000 rows . i am using for loop in my code , i think because of this my code execution time increased. if anyone know how to change this piece of code in to pandas package that reduce execution time
for i in range(len(df)):
if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('EOG')
df['EOG_flag'].values[i]=1
if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('gust')
df['Gust_flag'].values[i]=1
Note: this code is working well , just execution time is high
python-3.x pandas
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
I have 12000 csv files every file have 6000 rows . i am using for loop in my code , i think because of this my code execution time increased. if anyone know how to change this piece of code in to pandas package that reduce execution time
for i in range(len(df)):
if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('EOG')
df['EOG_flag'].values[i]=1
if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
#print('gust')
df['Gust_flag'].values[i]=1
Note: this code is working well , just execution time is high
python-3.x pandas
python-3.x pandas
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
asked Mar 28 at 3:47
NickelNickel
1398 bronze badges
1398 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
You can use vectorized solution - craete boolean mask separately, chain together by & and set values in numpy.where:
x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
m3 = df['Avg'].values > 2
m23 = m2 & m3
df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
Performance:
np.random.seed(2019)
N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)
In [91]: %%timeit
...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
...: m3 = df['Avg'].values > 2
...: m23 = m2 & m3
...:
...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
...:
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [93]: %%timeit
...: for i in range(len(df)):
...: if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('EOG')
...: df['EOG_flag'].values[i]=1
...:
...: if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('gust')
...: df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You can use vectorized solution - craete boolean mask separately, chain together by & and set values in numpy.where:
x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
m3 = df['Avg'].values > 2
m23 = m2 & m3
df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
Performance:
np.random.seed(2019)
N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)
In [91]: %%timeit
...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
...: m3 = df['Avg'].values > 2
...: m23 = m2 & m3
...:
...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
...:
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [93]: %%timeit
...: for i in range(len(df)):
...: if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('EOG')
...: df['EOG_flag'].values[i]=1
...:
...: if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('gust')
...: df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
add a comment |
3
You can use vectorized solution - craete boolean mask separately, chain together by & and set values in numpy.where:
x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
m3 = df['Avg'].values > 2
m23 = m2 & m3
df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
Performance:
np.random.seed(2019)
N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)
In [91]: %%timeit
...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
...: m3 = df['Avg'].values > 2
...: m23 = m2 & m3
...:
...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
...:
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [93]: %%timeit
...: for i in range(len(df)):
...: if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('EOG')
...: df['EOG_flag'].values[i]=1
...:
...: if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('gust')
...: df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
add a comment |
3
3
3
You can use vectorized solution - craete boolean mask separately, chain together by & and set values in numpy.where:
x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
m3 = df['Avg'].values > 2
m23 = m2 & m3
df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
Performance:
np.random.seed(2019)
N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)
In [91]: %%timeit
...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
...: m3 = df['Avg'].values > 2
...: m23 = m2 & m3
...:
...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
...:
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [93]: %%timeit
...: for i in range(len(df)):
...: if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('EOG')
...: df['EOG_flag'].values[i]=1
...:
...: if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('gust')
...: df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
You can use vectorized solution - craete boolean mask separately, chain together by & and set values in numpy.where:
x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
m3 = df['Avg'].values > 2
m23 = m2 & m3
df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
Performance:
np.random.seed(2019)
N = 6000
c = ['EOG_Start_model','EOG_Min_model','EOG_start_farm','EOG_Min_Farm','EOG_Max_model',
'EOG_Max_Farm','Avg','EOG_flag','Gust_flag']
df = pd.DataFrame(np.random.rand(N, 9), columns=c)
print (df)
In [91]: %%timeit
...: x = df['EOG_start_farm'].values-df['EOG_Min_Farm'].values
...: m1 = (df['EOG_Start_model'].values-df['EOG_Min_model'].values) < x
...: m2 = (df['EOG_Max_model'].values-df['EOG_Min_model'].values) < x
...: m3 = df['Avg'].values > 2
...: m23 = m2 & m3
...:
...: df['EOG_flag'] = np.where(m1 & m2 & m3, 1, df['EOG_flag'].values)
...: df['Gust_flag'] = np.where(m2 & m3, 1, df['Gust_flag'].values)
...:
597 µs ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [93]: %%timeit
...: for i in range(len(df)):
...: if ((df['EOG_Start_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_start_farm'].values[i]-df['EOG_Min_Farm'].values[i])) &((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('EOG')
...: df['EOG_flag'].values[i]=1
...:
...: if ((df['EOG_Max_model'].values[i]-df['EOG_Min_model'].values[i])<(df['EOG_Max_Farm'].values[i]-df['EOG_Min_Farm'].values[i]))&((df['Avg'].values[i]>2)):
...: #print('gust')
...: df['Gust_flag'].values[i]=1
231 ms ± 1.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
edited Mar 28 at 8:01
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
answered Mar 28 at 6:46
jezraeljezrael
406k32 gold badges423 silver badges486 bronze badges
406k32 gold badges423 silver badges486 bronze badges
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
add a comment |
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
this code have same result as for loop , there is no noticeable execution time,almost same execution time
– Nickel
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - OK, add some tests.
– jezrael
Mar 28 at 7:13
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - Improve solution and added performance timings.
– jezrael
Mar 28 at 8:12
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
@Nickel - It is 387 times faster like original solution.
– jezrael
Mar 28 at 8:13
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