在python中比较1d和2d系列的有效方法

我有一个数据帧，其中一列作为字符串数组，第二列作为一个字符串值

a = pd.Series([["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"]])

b = pd.Series(["a","d","e", "c", "b"])

我想检查a中是否包含b，但在运行isin函数时收到一个错误

b.isin(a)

任何解决办法。我在这里特别尝试避免循环，不确定这是否是一个基于运行时的好策略

编辑：

**a                          b**
["a","b","c", "d"]           a
["a","b","c", "d"]           d
["a","b","c", "d"]           e
["a","b","c", "d"]           c
["a","b","c", "d"]           b

预期输出是进行行比较的系列

[True True False True True]

---

这应该适合您：

import pandas as pd
a = pd.Series([["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"]])
b = pd.Series(["a","d","e", "c", "b"])
[x in y for y,x in zip(a,b)]

。。。其输出为：

[True, True, False, True, True]

---

pandas.Series

实现了一种方法，您可以通过以下方式使用该方法来查找

b

中也出现在

a

列表中的元素：

import pandas as pd

a = pd.Series([["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"],["a","b","c", "d"]])
b = pd.Series(["a","d","e", "c", "b"])

a.combine(b, lambda a,b: b in a)

输出：

0     True
1     True
2    False
3     True
4     True
dtype: object

---

预期的输出是什么？我不确定效率，但是

any（map（lambda x:x==b.tolist（），a））

应该可以工作；我编辑了帖子以包含所需的输出。感谢您指出这一点。

b

在序列中有5个字符串，但是

a

中的每一行只有4个元素。而且，

a

中的所有行都相同。你能说说为什么你希望

b

匹配前两行和后两行，而不是中间一行吗？啊！我明白你现在用最新的编辑是什么意思了。谢谢。谢谢分享一个不使用循环的方法。