使用Python中的Pandas，仅选择group by group count为1的行

我已按此处建议筛选了我的数据：

现在，我只想让作者在这个数据框中出现一次。我写了这封信，但不起作用：

def where_just_one_exists(group):
        return group.loc[group.count() == 1]
most_expensive_single_category = most_expensive_for_each_model.groupby('author', as_index = False).apply(where_just_one_exists).reset_index(drop = True)
print most_expensive_single_category

错误：

  File "/home/mike/anaconda/lib/python2.7/site-packages/pandas/core/indexing.py", line 1659, in check_bool_indexer
    raise IndexingError('Unalignable boolean Series key provided')
pandas.core.indexing.IndexingError: Unalignable boolean Series key provided

我期望的输出是：

    author        cat  val
0  author1  category2   15
1  author2  category4    9
2  author3  category1    7
3  author3  category3    7 

容易的

---

我的解决方案有点复杂，但仍然有效

def groupbyOneOccurrence(df):
    grouped = df.groupby("author")
    retDf = pd.DataFrame()
    for group in grouped:
        if len(group[1]._get_values) == 1:
            retDf = pd.concat([retDf, group[1]])
    return retDf


author        cat val
0  author1  category2  15
1  author2  category4   9

---

你想要的输出是什么？我已经添加了想要的输出。Padraic的解决方案似乎正是下面有人建议的。在应用count和mean之后，我将如何对其进行排序？

df.groupby('author').filter(lambda x: len(x)==1)


     author        cat  val
id                         
0   author1  category2   15
1   author2  category4    9

def groupbyOneOccurrence(df):
    grouped = df.groupby("author")
    retDf = pd.DataFrame()
    for group in grouped:
        if len(group[1]._get_values) == 1:
            retDf = pd.concat([retDf, group[1]])
    return retDf


author        cat val
0  author1  category2  15
1  author2  category4   9