如何检查列表中的所有元素是否都存在于熊猫列中

Question

我有一个数据框和一个列表：

df = pd.DataFrame({'id':[1,2,3,4,5,6,7,8], 
    'char':[['a','b'],['a','b','c'],['a','c'],['b','c'],[],['c','a','d'],['c','d'],['a']]})

names = ['a','c']

只有当a 和c 都出现在char 列中时，我才想获取行。（这里的顺序无关紧要）

预期输出：

       char  id                                                                                                                      
1  [a, b, c]   2                                                                                                                      
2     [a, c]   3                                                                                                                      
5  [c, a, d]   6   

我的努力

true_indices = []
for idx, row in df.iterrows():
    if all(name in row['char'] for name in names):
        true_indices.append(idx)


ids = df[df.index.isin(true_indices)]

这给了我正确的输出，但对于大型数据集来说太慢了，所以我正在寻找更有效的解决方案。

Answer 1

使用pd.DataFrame.apply:

df[df['char'].apply(lambda x: set(names).issubset(x))]

输出：

   id       char
1   2  [a, b, c]
2   3     [a, c]
5   6  [c, a, d]

Answer 2

您可以从名称列表中构建一个集合以便更快地查找，并使用set.issubset 检查集合中的所有元素是否包含在列列表中：

names = set(['a','c'])
df[df['char'].map(names.issubset)]

   id       char
1   2  [a, b, c]
2   3     [a, c]
5   6  [c, a, d]

Answer 3

对issubset使用列表推导：

mask = [set(names).issubset(x) for x in df['char']]
df = df[mask]
print (df)
   id       char
1   2  [a, b, c]
2   3     [a, c]
5   6  [c, a, d]

Series.map 的另一个解决方案：

df = df[df['char'].map(set(names).issubset)]
print (df)
   id       char
1   2  [a, b, c]
2   3     [a, c]
5   6  [c, a, d]

性能取决于行数和匹配值的数量：

df = pd.concat([df] * 10000, ignore_index=True)

In [270]: %timeit df[df['char'].apply(lambda x: set(names).issubset(x))]
45.9 ms ± 2.26 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [271]: %%timeit
     ...: names = set(['a','c'])
     ...: [names.issubset(set(row)) for _,row in df.char.iteritems()]
     ...: 
46.7 ms ± 5.51 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)


In [272]: %%timeit
     ...: df[[set(names).issubset(x) for x in df['char']]]
     ...: 
45.6 ms ± 1.26 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [273]: %%timeit
     ...: df[df['char'].map(set(names).issubset)]
     ...: 
18.3 ms ± 2.96 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [274]: %%timeit
     ...: n = set(names)
     ...: df[df['char'].map(n.issubset)]
     ...: 
16.6 ms ± 278 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [279]: %%timeit
     ...: names = set(['a','c'])
     ...: m = [name.issubset(i) for i in df.char.values.tolist()]
     ...: 
19.2 ms ± 317 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 4

试试这个。

df['char']=df['char'].apply(lambda x: x if ("a"in x and "c" in x) else np.nan)
print(df.dropna())

输出：

   id       char
1   2  [a, b, c]
2   3     [a, c]
5   6  [c, a, d]