from numpy import *list = array([["A", 25, 2], ["B", 25, 3], ["C", 10, 1], ["D", 50, 25]])如何按第二個元素除以第三個元素的降序大小對該數組進行排序。所以在這種情況下，正確的安排是：[“A”, 25, 2], [“C”, 10, 1], [“B”, 25, 3], [“D”, 50, 25]
2 回答
胡子哥哥
TA貢獻1825條經驗 獲得超6個贊
import numpy as np
# Let's use a as the variable name so we don't override the list keyword
a = np.array([["A", 25, 2], ["B", 25, 3], ["C", 10, 1], ["D", 50, 25]])
# Need to convert values to int, because they are being casted as string
# in the original array (arrays only support one data type)
key_values = a[:,1].astype(int) / a[:,2].astype(int)
# Use argsort to get a sort index, then reverse the order to make it descending
index_array = np.argsort(key_values)[::-1] # -1 reverses the order
print(a[index_array])
輸出：
[['A' '25' '2']
['C' '10' '1']
['B' '25' '3']
['D' '50' '25']]
隔江千里
TA貢獻1906條經驗 獲得超10個贊
你能用熊貓嗎？如果是這樣，
import numpy as np
import pandas as pd
df = pd.DataFrame(np.array([["A", 25, 2], ["B", 25, 3], ["C", 10, 1], ["D", 50, 25]]))
df[1] = pd.to_numeric(df[1])
df[2] = pd.to_numeric(df[2])
df[3] = df[1] / df[2]
sorted_list = df.sort_values(by=3, ascending=False).values[:,:3]
print(sorted_list)
array([['A', 25, 2],
['C', 10, 1],
['B', 25, 3],
['D', 50, 25]], dtype=object)
請注意，這里我假設（基于我對問題的理解）您想要根據第二列除以第三列的值進行排序。如果是這種情況，輸出示例的前兩個元素將被翻轉（因為 25/2 > 10/1）。
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