我正在尋找一種返回 (nxn) 數(shù)據(jù)幀的方法，其中數(shù)據(jù)幀的每個(gè)值都是兩個(gè)數(shù)據(jù)幀的值之間的交集數(shù)（兩者的大小均為 nxn）。我不太確定如何在兩個(gè)數(shù)據(jù)幀之間進(jìn)行這樣的操作。任何幫助是極大的贊賞。謝謝！df1 0 10 [4, 7, 3, 5] [4, 7, 3, 5]1 [8, 2, 6] [8, 2, 6]2 [9, 1, 8, 2] [9, 1, 8, 2]3 [3, 5] [3, 5]4 [9, 4, 8] [9, 4, 8]5 [0, 1, 4] [0, 1, 4]df2 0 10 [2, 3, 6, 9] [6, 2, 3, 5]1 [2, 3, 6, 9] [6, 2, 3, 5]2 [2, 3, 6, 9] [6, 2, 3, 5]3 [2, 3, 6, 9] [6, 2, 3, 5]4 [2, 3, 6, 9] [6, 2, 3, 5]5 [2, 3, 6, 9] [6, 2, 3, 5]df3 - intended dataframe to be returned 0 10 1 21 1 22 2 13 1 24 0 05 0 0編輯：修復(fù)了示例結(jié)果的錯(cuò)誤
1 回答
慕碼人2483693
TA貢獻(xiàn)1860條經(jīng)驗(yàn) 獲得超9個(gè)贊
嗯，我無法直接對(duì) pandas 執(zhí)行此操作，我只有字典的解決方案。我認(rèn)為您的示例結(jié)果有錯(cuò)誤，我認(rèn)為我的結(jié)果是預(yù)期的結(jié)果。
import pandas as pd
# Prework to get your data
data = {0: [[4, 7, 3, 5], [8, 2, 6], [9, 1, 8, 2], [3, 5], [9, 4, 8], [0, 1, 4]],
1: [[4, 7, 3, 5], [8, 2, 6], [9, 1, 8, 2], [3, 5], [9, 4, 8], [0, 1, 4]]}
data2 = {0: [[2, 3, 6, 9], [2, 3, 6, 9], [2, 3, 6, 9], [2, 3, 6, 9], [2, 3, 6, 9], [2, 3, 6, 9]],
1: [[6, 2, 3, 5], [6, 2, 3, 5], [6, 2, 3, 5], [6, 2, 3, 5], [6, 2, 3, 5], [6, 2, 3, 5]]}
df = pd.DataFrame(data)
df2 = pd.DataFrame(data2)
# ---
dc = df.to_dict()
dc2 = df2.to_dict()
new_dc = dc.copy()
for key in dc:
for val in dc[key]:
new_dc[key][val] = len(set(dc[key][val]).intersection(dc2[key][val]))
new_df = pd.DataFrame(new_dc)
print(new_df)
輸出：
0 1
0 1 2
1 2 2
2 2 1
3 1 2
4 1 0
5 0 0
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