我想模擬不同天數的個體生長和死亡率變化。我的資料框格式如下...
import pandas as pd
data = {'unique_id': ['2', '4', '5', '13'],
'length': ['27.7', '30.2', '25.4', '29.1'],
'no_fish': ['3195', '1894', '8', '2774'],
'days_left': ['253', '253', '254', '256'],
'growth': ['0.3898', '0.3414', '0.4080', '0.3839']
}
df = pd.DataFrame(data)
print(df)
unique_id length no_fish days_left growth
0 2 27.7 3195 253 0.3898
1 4 30.2 1894 253 0.3414
2 5 25.4 8 254 0.4080
3 13 29.1 2774 256 0.3839
理想情況下，我希望初始長度（即長度）增加一年中剩余的每一天（即 days_left）的每日增長率（即增長）。
df['final'] = df['length'] (df['days_left'] * df['growth']
但是，我還想使用特定于大小的方程每天更新每個人所代表的魚的數量（即 no_fish）。我對python相當陌生，所以我最初想使用for回圈（我不確定是否有另一種更有效的方法）。我的代碼如下：
# keep track of run time - START
start_time = time.perf_counter()
df['z'] = 0.0
for indx in range(len(df)):
count = 1
while count <= int(df.days_to_forecast[indx]):
# (1) update individual length
df.lgth[indx] = df.lgth[indx] df.linearGR[indx]
# (2) estimate daily size-specific mortality
if df.lgth[indx] > 50.0:
df.z[indx] = 0.01
else:
if df.lgth[indx] <= 50.0:
df.z[indx] = 0.052857-((0.03/35)*df.lgth[indx])
elif df.lgth[indx] < 15.0:
df.z[indx] = 0.728*math.exp(-0.1892*df.lgth[indx])
df['no_fish'].round(decimals = 0)
if df.no_fish[indx] < 1.0:
df.no_fish[indx] = 0.0
elif df.no_fish[indx] >= 1.0:
df.no_fish[indx] = df.no_fish[indx]*math.exp(-(df.z[indx]))
# (3) reduce no. of days left in forecast by 1
count = count 1
# keep track of run time - END
total_elapsed_time = round(time.perf_counter() - start_time, 2)
print("Forecast iteration completed in {} seconds".format(total_elapsed_time))
上面的代碼現在可以正常作業了，但是對于 40,000 個人每人運行 200 多天仍然是低效的。
我非常感謝有關如何修改以下代碼以使其成為 pythonic 的任何建議。
謝謝
uj5u.com熱心網友回復：
向我建議的另一個選項是使用pd.dataframe.apply函式。這極大地減少了整體運行時間，并且將來可能對其他人有用。
### === RUN SIMULATION === ###
start_time = time.perf_counter() # keep track of run time -- START
#-------------------------------------------------------------------------#
def function_to_apply( df ):
df['z_instantMort'] = ''
for indx in range(int(df['days_left'])):
# (1) update individual length
df['length'] = df['length'] df['growth']
# (2) estimate daily size-specific mortality
if df['length'] > 50.0:
df['z_instantMort'] = 0.01
else:
if df['length'] <= 50.0:
df['z_instantMort'] = 0.052857-((0.03/35)*df['length'])
elif df['length'] < 15.0:
df['z_instantMort'] = 0.728*np.exp(-0.1892*df['length'])
whole_fish = round(df['no_fish'], 0)
if whole_fish < 1.0:
df['no_fish'] = 0.0
elif whole_fish >= 1.0:
df['no_fish'] = df['no_fish']*np.exp(-(df['z_instantMort']))
return df
#-------------------------------------------------------------------------#
sim_results = df.apply(function_to_apply, axis=1)
total_elapsed_time = round(time.perf_counter() - start_time, 2) # END
print("Forecast iteration completed in {} seconds".format(total_elapsed_time))
print(sim_results)
### ====================== ###
輸出是...
Forecast iteration completed in 0.05 seconds
unique_id length no_fish days_left growth z_instantMort
0 2.0 126.3194 148.729190 253.0 0.3898 0.01
1 4.0 116.5742 93.018465 253.0 0.3414 0.01
2 5.0 129.0320 0.000000 254.0 0.4080 0.01
3 13.0 127.3784 132.864757 256.0 0.3839 0.01
uj5u.com熱心網友回復：
正如我在評論中所說，在此設定中 for 回圈的一個更可取的替代方法是使用向量運算。例如，運行您的代碼：
import pandas as pd
import time
import math
import numpy as np
data = {'unique_id': [2, 4, 5, 13],
'length': [27.7, 30.2, 25.4, 29.1],
'no_fish': [3195, 1894, 8, 2774],
'days_left': [253, 253, 254, 256],
'growth': [0.3898, 0.3414, 0.4080, 0.3839]
}
df = pd.DataFrame(data)
print(df)
# keep track of run time - START
start_time = time.perf_counter()
df['z'] = 0.0
for indx in range(len(df)):
count = 1
while count <= int(df.days_left[indx]):
# (1) update individual length
df.length[indx] = df.length[indx] df.growth[indx]
# (2) estimate daily size-specific mortality
if df.length[indx] > 50.0:
df.z[indx] = 0.01
else:
if df.length[indx] <= 50.0:
df.z[indx] = 0.052857-((0.03/35)*df.length[indx])
elif df.length[indx] < 15.0:
df.z[indx] = 0.728*math.exp(-0.1892*df.length[indx])
df['no_fish'].round(decimals = 0)
if df.no_fish[indx] < 1.0:
df.no_fish[indx] = 0.0
elif df.no_fish[indx] >= 1.0:
df.no_fish[indx] = df.no_fish[indx]*math.exp(-(df.z[indx]))
# (3) reduce no. of days left in forecast by 1
count = count 1
# keep track of run time - END
total_elapsed_time = round(time.perf_counter() - start_time, 2)
print("Forecast iteration completed in {} seconds".format(total_elapsed_time))
print(df)
輸出：
unique_id length no_fish days_left growth
0 2 27.7 3195 253 0.3898
1 4 30.2 1894 253 0.3414
2 5 25.4 8 254 0.4080
3 13 29.1 2774 256 0.3839
Forecast iteration completed in 31.75 seconds
unique_id length no_fish days_left growth z
0 2 126.3194 148.729190 253 0.3898 0.01
1 4 116.5742 93.018465 253 0.3414 0.01
2 5 129.0320 0.000000 254 0.4080 0.01
3 13 127.3784 132.864757 256 0.3839 0.01
現在使用矢量操作，您可以執行以下操作：
# keep track of run time - START
start_time = time.perf_counter()
df['z'] = 0.0
for day in range(1, df.days_left.max() 1):
update = day <= df['days_left']
# (1) update individual length
df[update]['length'] = df[update]['length'] df[update]['growth']
# (2) estimate daily size-specific mortality
df[update]['z'] = np.where( df[update]['length'] > 50.0, 0.01, 0.052857-( ( 0.03 / 35)*df[update]['length'] ) )
df[update]['z'] = np.where( df[update]['length'] < 15.0, 0.728 * np.exp(-0.1892*df[update]['length'] ), df[update]['z'] )
df[update]['no_fish'].round(decimals = 0)
df[update]['no_fish'] = np.where(df[update]['no_fish'] < 1.0, 0.0, df[update]['no_fish'] * np.exp(-(df[update]['z'])))
# keep track of run time - END
total_elapsed_time = round(time.perf_counter() - start_time, 2)
print("Forecast iteration completed in {} seconds".format(total_elapsed_time))
print(df)
帶輸出
Forecast iteration completed in 1.32 seconds
unique_id length no_fish days_left growth z
0 2 126.3194 148.729190 253 0.3898 0.0
1 4 116.5742 93.018465 253 0.3414 0.0
2 5 129.0320 0.000000 254 0.4080 0.0
3 13 127.3784 132.864757 256 0.3839 0.0
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