我有這個 df （帶有示例所需的結(jié)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"]]})我將（到目前為止）字符串“翻譯”為新含義并存儲為元素列表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，我想通過外部列表提取一些翻譯后的country_name系列，american并將它們放在一個單獨的系列中（all_but_american）american = ["United States", "Brazil"]結(jié)果應(yīng)該與all_but_american_desired系列相同。到目前為止我嘗試過的：dfn["all_but_american1"] = dfn["country_name"].apply(lambda x: [i for i in x if i not in american])我之前使用過嘗試1非常相同的方法并且它有效，但是這次沒有任何效果，我找不到原因（這次我也嘗試過其他方法，但由于我對它們不熟悉）不會發(fā)帖）...有人可以檢查一下嗎？如果可能的話，也解釋一下我做錯了什么。
1 回答
MM們
TA貢獻1886條經(jīng)驗 獲得超2個贊
要country_name創(chuàng)建列表，而不是使用連接值的元素列表：
dfn["country_name"] = dfn["country_code"].apply(lambda x: [masked[i] for i in x.split(", ")])
然后你的第二個解決方案運行良好：
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]
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