[英]Compute if value exists in a column on lists in pandas dataframe
我的數據框中有2列
類似客戶購買的產品ID列表“p_list”
df = pd.DataFrame({'p': [12, 4, 5, 6, 7, 7, 6,5],'p_list':[[12,1,5], [3,1],[8,9,11], [6,7,9], [7,1,2],[12,9,8], [6,1,15],[6,8,9,11]]})
我想檢查“p_list”上是否存在“p”,所以我應用了這段代碼
df["exist"]= df.apply(lambda r: 1 if r["p"] in r["p_list"] else 0, axis=1)
問題是我在這個數據幀中有大約5000萬行,因此執行需要很長時間。
是否有更有效的方法來計算此列?
謝謝。
您可以使用list comprehension
,最后將True, False
值轉換為int
:
df["exist"] = [r[0] in r[1] for r in zip(df["p"], df["p_list"])]
df["exist"] = df["exist"].astype(int)
print (df)
p p_list exist
0 12 [12, 1, 5] 1
1 4 [3, 1] 0
2 5 [8, 9, 11] 0
3 6 [6, 7, 9] 1
4 7 [7, 1, 2] 1
5 7 [12, 9, 8] 0
6 6 [6, 1, 15] 1
7 5 [6, 8, 9, 11] 0
df["exist"] = [int(r[0] in r[1]) for r in zip(df["p"], df["p_list"])]
print (df)
p p_list exist
0 12 [12, 1, 5] 1
1 4 [3, 1] 0
2 5 [8, 9, 11] 0
3 6 [6, 7, 9] 1
4 7 [7, 1, 2] 1
5 7 [12, 9, 8] 0
6 6 [6, 1, 15] 1
7 5 [6, 8, 9, 11] 0
時間 :
#[8000 rows x 2 columns]
df = pd.concat([df]*1000).reset_index(drop=True)
print (df)
In [89]: %%timeit
...: df["exist2"] = [r[0] in r[1] for r in zip(df["p"], df["p_list"])]
...: df["exist2"] = df["exist2"].astype(int)
...:
100 loops, best of 3: 6.07 ms per loop
In [90]: %%timeit
...: df["exist"] = [1 if r[0] in r[1] else 0 for r in zip(df["p"], df["p_list"])]
...:
100 loops, best of 3: 7.16 ms per loop
In [91]: %%timeit
...: df["exist"] = [int(r[0] in r[1]) for r in zip(df["p"], df["p_list"])]
...:
100 loops, best of 3: 9.23 ms per loop
In [92]: %%timeit
...: df['exist1'] = df.apply(lambda x: x.p in x.p_list, axis=1).astype(int)
...:
1 loop, best of 3: 370 ms per loop
In [93]: %%timeit
...: df["exist"]= df.apply(lambda r: 1 if r["p"] in r["p_list"] else 0, axis=1)
1 loop, best of 3: 310 ms per loop
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