Python 如何确定 Pandas 列是否包含特定值

How to determine whether a Pandas Column contains a particular value

提问by Michael

I am trying to determine whether there is an entry in a Pandas column that has a particular value. I tried to do this with if x in df['id']. I thought this was working, except when I fed it a value that I knew was not in the column 43 in df['id']it still returned True. When I subset to a data frame only containing entries matching the missing id df[df['id'] == 43]there are, obviously, no entries in it. How to I determine if a column in a Pandas data frame contains a particular value and why doesn't my current method work? (FYI, I have the same problem when I use the implementation in this answerto a similar question).

我正在尝试确定 Pandas 列中是否有具有特定值的条目。我试图用if x in df['id']. 我认为这是有效的，除非我给它提供了一个我知道不在列中的值，43 in df['id']但它仍然返回True。当我将数据框子集化为仅包含与缺少的 id 匹配的条目时df[df['id'] == 43]，显然其中没有条目。如何确定 Pandas 数据框中的列是否包含特定值，为什么我当前的方法不起作用？（仅供参考，当我在这个类似问题的答案中使用实现时，我遇到了同样的问题）。

采纳答案by Andy Hayden

inof a Series checks whether the value is in the index:

in系列的检查值是否在索引中：

In [11]: s = pd.Series(list('abc'))

In [12]: s
Out[12]: 
0    a
1    b
2    c
dtype: object

In [13]: 1 in s
Out[13]: True

In [14]: 'a' in s
Out[14]: False

One option is to see if it's in uniquevalues:

一种选择是查看它是否具有唯一值：

In [21]: s.unique()
Out[21]: array(['a', 'b', 'c'], dtype=object)

In [22]: 'a' in s.unique()
Out[22]: True

or a python set:

或 python 集：

In [23]: set(s)
Out[23]: {'a', 'b', 'c'}

In [24]: 'a' in set(s)
Out[24]: True

As pointed out by @DSM, it may be more efficient (especially if you're just doing this for one value) to just use in directly on the values:

正如@DSM 所指出的，直接在值上使用 in 可能更有效（特别是如果您只是为一个值执行此操作）：

In [31]: s.values
Out[31]: array(['a', 'b', 'c'], dtype=object)

In [32]: 'a' in s.values
Out[32]: True

回答by ffeast

You can also use pandas.Series.isinalthough it's a little bit longer than 'a' in s.values:

您也可以使用pandas.Series.isin虽然它比'a' in s.values以下长一点：

In [2]: s = pd.Series(list('abc'))

In [3]: s
Out[3]: 
0    a
1    b
2    c
dtype: object

In [3]: s.isin(['a'])
Out[3]: 
0    True
1    False
2    False
dtype: bool

In [4]: s[s.isin(['a'])].empty
Out[4]: False

In [5]: s[s.isin(['z'])].empty
Out[5]: True

But this approach can be more flexible if you need to match multiple values at once for a DataFrame (see DataFrame.isin)

但是，如果您需要为 DataFrame 一次匹配多个值，则这种方法会更加灵活（请参阅DataFrame.isin）

>>> df = DataFrame({'A': [1, 2, 3], 'B': [1, 4, 7]})
>>> df.isin({'A': [1, 3], 'B': [4, 7, 12]})
       A      B
0   True  False  # Note that B didn't match 1 here.
1  False   True
2   True   True

回答by Eli B

Simple condition:

简单条件：

if any(str(elem) in ['a','b'] for elem in df['column'].tolist()):

回答by U10-Forward

Or use Series.tolistor Series.any:

或使用Series.tolist或Series.any：

>>> s = pd.Series(list('abc'))
>>> s
0    a
1    b
2    c
dtype: object
>>> 'a' in s.tolist()
True
>>> (s=='a').any()
True

Series.tolistmakes a list about of a Series, and the other one i am just getting a boolean Seriesfrom a regular Series, then checking if there are any Trues in the boolean Series.

Series.tolist制作一个关于 a 的列表Series，另一个我只是Series从常规中获取一个布尔值Series，然后检查布尔值中是否有任何Trues Series。

回答by Shahir Ansari

found = df[df['Column'].str.contains('Text_to_search')]
print(found.count())

the found.count()will contains number of matches

在found.count()遗嘱中含有的比赛数量

And if it is 0 then means string was not found in the Column.

如果它是 0 则表示在列中找不到字符串。

回答by Vicky Ding

I don't suggest to use "value in series", which can lead many errors. Please see this answer for detail: Using in operator with Pandas series

我不建议使用“串联值”，这会导致很多错误。有关详细信息，请参阅此答案：Using in operator with Pandas series

回答by Allen Wang

I did a few simple tests:

我做了一些简单的测试：

In [10]: x = pd.Series(range(1000000))

In [13]: timeit 999999 in x.values
567 μs ± 25.6 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [15]: timeit x.isin([999999]).any()
9.54 ms ± 291 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [16]: timeit (x == 999999).any()
6.86 ms ± 107 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [17]: timeit 999999 in set(x)
79.8 ms ± 1.98 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [21]: timeit x.eq(999999).any()
7.03 ms ± 33.7 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [22]: timeit x.eq(9).any()
7.04 ms ± 60 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [24]: timeit 9 in x.values
666 μs ± 15.7 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Interestingly it doesn't matter if you look up 9 or 999999, it seems like it takes about the same amount of time using the in syntax (must be using binary search)

有趣的是，查找 9 或 999999 并不重要，使用 in 语法似乎花费的时间大致相同（必须使用二进制搜索）

In [24]: timeit 9 in x.values
666 μs ± 15.7 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [25]: timeit 9999 in x.values
647 μs ± 5.21 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [26]: timeit 999999 in x.values
642 μs ± 2.11 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [27]: timeit 99199 in x.values
644 μs ± 5.31 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [28]: timeit 1 in x.values
667 μs ± 20.8 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Seems like using x.values is the fastest, but maybe there is a more elegant way in pandas?

似乎使用 x.values 是最快的，但也许在大熊猫中有更优雅的方式？

回答by Ramana Sriwidya

Use

用

df[df['id']==x].index.tolist()

If xis present in idthen it'll return the list of indices where it is present, else it gives an empty list.

如果x存在，id那么它将返回它存在的索引列表，否则它给出一个空列表。

回答by Namrata Tolani

Suppose you dataframe looks like :

假设你的数据框看起来像：

Now you want to check if filename "80900026941984" is present in the dataframe or not.

现在您要检查数据框中是否存在文件名“80900026941984”。

You can simply write :

你可以简单地写：

if sum(df["filename"].astype("str").str.contains("80900026941984")) > 0:
    print("found")