Question

我有一个包含电影类型和ID的列表，还有第二个仅包含ID的列表。现在，我尝试使用第一个列表仅获取列表中的流派。这个问题应该很容易，但是我找不到解决方案。

带有电影类型的列表是通过csv文件加载的。看起来像：

   Unnamed: 0     id                name
0            0     28              Action
1            1     12           Adventure
2            2     16           Animation
3            3     35              Comedy
4            4     80               Crime
5            5     99         Documentary
6            6     18               Drama
7            7  10751              Family
8            8     14             Fantasy
9            9     36             History
10          10     27              Horror

第二个列表如下：

 filmgenre = [28, 16, 14, 99]

到目前为止，我的代码：

filmgenreslist = pd.read_csv("genrelist.csv") #load csv
cleanfilmlist = filmgenreslist.loc[:,"name"] #tried to reduce the list to necessary informations
filmgenre = [28, 16, 14, 878]
filmgenrenames = [] #tried to save the genres for later
number = 0 #just to test
test = cleanfilmlist.loc[filmgenreslist["id"] == filmgenre[number]] #try to extract only the genrename
print(test)
#print(test[-1]) 
#to loop through the data, multiple tries but can't solve
for number in range(5):
    try:
        #filmgenrenames.append([cleanfilmlist["name"].where(cleanfilmlist["id"] == filmgenre[number])])
        filmgenrenames.append([cleanfilmlist.loc(cleanfilmlist["id"] == filmgenre[number])])
        #filmgenrenames.append([cleanfilmlist.loc[filmgenreslist["id"] == filmgenre[number]]])
        #print(number, filmgenrenames)
    except:
        filmgenrenames.append(["missing"])

print(filmgenrenames)

我很累得到如下输出：

filmgenresname =["Action","Animation","Fantasy","Documentary","missing"]

Answer 1

您可以将id作为键的字典，并将name作为值的字典，然后将其与列表映射：

map_id_name = dict(zip(filmgenreslist['id'], filmgenreslist['name']))
list(map(lambda x:map_id_name.get(x, "missing"), filmgenre ))

熊猫从列表中获得价值并创建新列表

1 个答案: