Python 如何从dataframe中的两列除法中找到最小值

我想找到两列的最小除法，只在列表的第三列中有值。 我的数据帧是：

   ID  size  price
0   1     5    300
1   2    10    500
2   3    20    600
3   4    35    800
4   5    65    900
5   6    70   1000 

我只想从列表中有值的ID中找到最低价格/尺寸

ids_wanted = [1,4,6]

我编写了这段代码，它是有效的，但我觉得为任务创建新的数据帧既昂贵又没有必要

import numpy as np
import pandas as pd
index = [0,1,2,3,4,5]
i = pd.Series([1,2,3,4,5,6], index=index)
s = pd.Series([5,10,20,35,65,70],index= index)
p = pd.Series([300,500,600,800,900,1000],index= index)
df = pd.DataFrame(np.c_[i,s,p],columns = ["ID","size","price"])
print("original df:\n",df,"\n")

ids_wanted = [1,4,6]
df_with_ids_wanted = df.loc[df['ID'].isin(ids_wanted)]
print("df with ids wanted:\n",df_with_ids_wanted,"\n")
price_per_byte = df_with_ids_wanted['price'] / df_with_ids_wanted['size']
df_with_ids_wanted_ppb = df_with_ids_wanted.assign(pricePerByte=price_per_byte)
print("df with ids wanted and price/size column:\n",df_with_ids_wanted_pps,"\n")
min_ppb = df_with_ids_wanted_pps['pricePerByte'].min()
print("min price per byte:",min_ppb)

产出：

original df:
    ID  size  price
0   1     5    300
1   2    10    500
2   3    20    600
3   4    35    800
4   5    65    900
5   6    70   1000 

df with ids wanted:
    ID  size  price
0   1     5    300
3   4    35    800
5   6    70   1000 

df with ids wanted and price/size column:
    ID  size  price  pricePerByte
0   1     5    300     60.000000
3   4    35    800     22.857143
5   6    70   1000     14.285714 

min price per byte: 14.285714285714286

---

如果您想简洁，可以尝试以下方法：

i = range(1,7)
s = [5,10,20,35,65,70]
p = [300,500,600,800,900,1000]
df = pd.DataFrame({"ID":i,"size":s,"price":p})
df

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

下一行将如下所示：

id_chosen = [1,4,6]
(df[df.ID.isin(id_chosen)]["price"]/df[df.ID.isin(id_chosen)]["size"]).min()

import numpy as np
import pandas as pd

dict = {'id': [1, 2, 3, 4, 5, 6],
        'size': [5, 10, 20, 35, 65, 70],
        'price': [300, 500, 600, 800, 900, 1000]
       }

df = pd.DataFrame(dict)

df['price/byte'] = df['price'] / df['size']

ids_wanted = [1, 4, 6]

subset = df[df['id'].isin(ids_wanted)]

sorted_values = subset.sort_values(by='price/byte', ascending = True)

print(sorted_values['price/byte'].iloc[0])

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

或

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

这样，您就不必创建新的数据帧。

---

希望这有帮助。

如果您想简洁，可以尝试以下方法：

i = range(1,7)
s = [5,10,20,35,65,70]
p = [300,500,600,800,900,1000]
df = pd.DataFrame({"ID":i,"size":s,"price":p})
df

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

下一行将如下所示：

id_chosen = [1,4,6]
(df[df.ID.isin(id_chosen)]["price"]/df[df.ID.isin(id_chosen)]["size"]).min()

import numpy as np
import pandas as pd

dict = {'id': [1, 2, 3, 4, 5, 6],
        'size': [5, 10, 20, 35, 65, 70],
        'price': [300, 500, 600, 800, 900, 1000]
       }

df = pd.DataFrame(dict)

df['price/byte'] = df['price'] / df['size']

ids_wanted = [1, 4, 6]

subset = df[df['id'].isin(ids_wanted)]

sorted_values = subset.sort_values(by='price/byte', ascending = True)

print(sorted_values['price/byte'].iloc[0])

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

或

输出：

    ID  size    price
0   1   5   300
1   2   10  500
2   3   20  600
3   4   35  800
4   5   65  900
5   6   70  1000

14.285714285714286

the minimum price/size is 14.285714285714286

这样，您就不必创建新的数据帧。

---

希望这有帮助。

我会这样做：

id_chosen = [1,4,6]
(df[df.ID.isin(id_chosen)]["price"]/df[df.ID.isin(id_chosen)]["size"]).min()

import numpy as np
import pandas as pd

dict = {'id': [1, 2, 3, 4, 5, 6],
        'size': [5, 10, 20, 35, 65, 70],
        'price': [300, 500, 600, 800, 900, 1000]
       }

df = pd.DataFrame(dict)

df['price/byte'] = df['price'] / df['size']

ids_wanted = [1, 4, 6]

subset = df[df['id'].isin(ids_wanted)]

sorted_values = subset.sort_values(by='price/byte', ascending = True)

print(sorted_values['price/byte'].iloc[0])

---

我会这样做：

id_chosen = [1,4,6]
(df[df.ID.isin(id_chosen)]["price"]/df[df.ID.isin(id_chosen)]["size"]).min()

import numpy as np
import pandas as pd

dict = {'id': [1, 2, 3, 4, 5, 6],
        'size': [5, 10, 20, 35, 65, 70],
        'price': [300, 500, 600, 800, 900, 1000]
       }

df = pd.DataFrame(dict)

df['price/byte'] = df['price'] / df['size']

ids_wanted = [1, 4, 6]

subset = df[df['id'].isin(ids_wanted)]

sorted_values = subset.sort_values(by='price/byte', ascending = True)

print(sorted_values['price/byte'].iloc[0])

---

我建议尽可能避免在pandas/NumPy中使用for循环。我建议尽可能避免pandas/NumPy中的for循环。更快的方法比比皆是