Python 如何检查列表中的所有元素是否都在列中
我有一个数据框和一个列表:Python 如何检查列表中的所有元素是否都在列中,python,python-3.x,pandas,Python,Python 3.x,Pandas,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'] 我只想在char列中同时存在a和c时获取行。(这里的顺序不重要) 预期输出: char id
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']
我只想在char
列中同时存在a
和c
时获取行。(这里的顺序不重要)
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速度太慢,因此我正在寻找更有效的解决方案。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]
将列表理解用于: 另一个解决方案包括: 性能取决于行数和匹配值数:
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)
您可以从名称列表中生成一个集合,以便更快地查找,并使用检查集合中的所有元素是否都包含在列列表中:
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]
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]
这个比其他的快。谢谢:-)@yatu-hmm,对我来说不是,但实际数据似乎不同
%%timeit names=set(['a','c'])m=[name.issubset(i)for i in df.char.values.tolist()]19.2ms±317µs/循环(平均±标准偏差为7次运行,每个循环100次)
df['char']=df['char'].apply(lambda x: x if ("a"in x and "c" in x) else np.nan)
print(df.dropna())
id char
1 2 [a, b, c]
2 3 [a, c]
5 6 [c, a, d]