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')

更新於: 14-Oct-2021

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