3 回答
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TA贡献1821条经验 获得超4个赞
您可以使用np.where():
import pandas as pd
df = pd.DataFrame(data={'hand' : ['left','left','both','both'],
'exp_force' : [25,28,82,84],
'left_max' : [38,38,38,38],
'both_max' : [90,90,90,90]})
df['ratio'] = np.where((df['hand']=='left'), df['exp_force'] / df['left_max'], df['exp_force'] / df['both_max'])
df
Out[42]:
hand exp_force left_max both_max ratio
0 left 25 38 90 0.657895
1 left 28 38 90 0.736842
2 both 82 38 90 0.911111
3 both 84 38 90 0.933333
或者,在现实生活中,如果您有很多条件和结果,那么您可以使用np.select(),这样您就不必np.where()像我在旧代码中所做的那样不断重复您的语句。最好np.select在这些情况下使用:
import pandas as pd
df = pd.DataFrame(data={'hand' : ['left','left','both','both'],
'exp_force' : [25,28,82,84],
'left_max' : [38,38,38,38],
'both_max' : [90,90,90,90]})
c1 = (df['hand']=='left')
c2 = (df['hand']=='both')
r1 = df['exp_force'] / df['left_max']
r2 = df['exp_force'] / df['both_max']
conditions = [c1,c2]
results = [r1,r2]
df['ratio'] = np.select(conditions,results)
df
Out[430]:
hand exp_force left_max both_max ratio
0 left 25 38 90 0.657895
1 left 28 38 90 0.736842
2 both 82 38 90 0.911111
3 both 84 38 90 0.933333
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TA贡献1827条经验 获得超9个赞
枚举
for i,e in enumerate(df['hand']):
if e == 'left':
df.at[i,'ratio'] = df.at[i,'exp_force'] / df.at[i,'left_max']
if e == 'both':
df.at[i,'ratio'] = df.at[i,'exp_force'] / df.at[i,'both_max']
df
输出:
hand exp_force left_max both_max ratio
0 left 25 38 90 0.657895
1 left 28 38 90 0.736842
2 both 82 38 90 0.911111
3 both 84 38 90 0.933333
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TA贡献1817条经验 获得超6个赞
您可以使用apply()数据框的方法:
df['ratio'] = df.apply(
lambda x: x['exp_force'] / x['left_max'] if x['hand']=='left' else x['exp_force'] / x['both_max'],
axis=1
)
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