Python Pandas - 返回不在其他索引中的索引的新索引,但不排序結果
要返回不在其他索引中的索引的新索引,但不排序結果,請使用 **difference()** 方法。將 **sort** 引數設定為 **False**。
首先,匯入所需的庫 −
import pandas as pd
建立兩個 Pandas 索引 −
index1 = pd.Index([30, 10, 20, 50, 40]) index2 = pd.Index([80, 40, 60, 20, 55])
顯示 Pandas index1 和 index2 −
print("Pandas Index1...\n",index1) print("Pandas Index2...\n",index2)
獲取兩個索引的差異。使用值為“False”的“sort”引數對結果進行不排序 −
res = index1.difference(index2, sort=False)
示例
以下為程式碼 −
import pandas as pd # Creating two Pandas index index1 = pd.Index([30, 10, 20, 50, 40]) index2 = pd.Index([80, 40, 60, 20, 55]) # Display the Pandas index1 and index2 print("Pandas Index1...\n",index1) print("Pandas Index2...\n",index2) # Return the number of elements in Index1 and Index2 print("\nNumber of elements in index1...\n",index1.size) print("\nNumber of elements in index2...\n",index2.size) # Get the difference of both the indexes # Results are unsorted using the "sort" parameter with value "False" res = index1.difference(index2, sort=False) # Difference of both the indexes i.e. return a new Index with elements of index not in other print("\nDifference...\n",res)
輸出
這將產生以下輸出 −
Pandas Index1... Int64Index([30, 10, 20, 50, 40], dtype='int64') Pandas Index2... Int64Index([80, 40, 60, 20, 55], dtype='int64') Number of elements in index1... 5 Number of elements in index2... 5 Difference... Int64Index([30, 10, 50], dtype='int64')
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