从 pandas 列表系列中提取元素并存储为单独的系列

Question

我有这个 df（带有样本期望结果）

dfn = pd.DataFrame({"country_code": ["USA, UK, FRA", "RUS, ZHC, JAP", "IN, BRA, ES"], 
                    "all_but_american_desired": [["United Kingdom", "France"], ["Russia", "China", "Japan"], ["India", "Spain"]]})

我将（到目前为止）字符串“翻译”成新的含义并存储为元素列表

masked = {"USA":"United States", "UK":"United Kingdom", "FRA":"France", 
          "RUS":"Russia", "ZHC":"China", "JAP":"Japan", 
          "IN":"India", "BRA":"Brazil", "ES":"Spain"}

dfn["country_name"] = dfn["country_code"].apply(lambda x: [", ".join({masked[i] for i in x.split(", ")})])

然后我想提取一些由外部列表翻译的 country_name 系列，american并将它们放在一个单独的系列中 (all_but_american)

american = ["United States", "Brazil"]

结果应与 all_but_american_desired 系列相同。到目前为止我尝试了什么：

dfn["all_but_american1"] = dfn["country_name"].apply(lambda x: [i for i in x if i not in american])

我之前用过 attempt1 完全相同的方法并且它有效，但是这次没有任何效果而且我找不到它的原因（这次我也尝试了其他方法，但是因为我不熟悉他们我会避免发布）......有人可以检查一下吗？如果可能的话，也解释一下我做错了什么。

Answer 1

对于 country_name 创建列表而不是一个具有连接值的元素列表：

dfn["country_name"] = dfn["country_code"].apply(lambda x: [masked[i] for i in x.split(", ")])

然后你的第二个解决方案运行良好：

american = ["United States", "Brazil"]

dfn["all_but_american1"] = dfn["country_name"].apply(lambda x: [i for i in x if i not in american])
print (dfn)
    country_code  all_but_american_desired  \
0   USA, UK, FRA  [United Kingdom, France]   
1  RUS, ZHC, JAP    [Russia, China, Japan]   
2    IN, BRA, ES            [India, Spain]   

                              country_name         all_but_american1  
0  [United States, United Kingdom, France]  [United Kingdom, France]  
1                   [Russia, China, Japan]    [Russia, China, Japan]  
2                   [India, Brazil, Spain]            [India, Spain]

从 pandas 列表系列中提取元素并存储为单独的系列

Extract element from pandas list-series and store as separate series

python

expand

extract

list

pandas