Question

我要列出100多个列表。我怎样才能只制造一个然后遍历它们？它比通常的字典循环更复杂。

list(i)=[]
for a in dict(i)['a']:
    d={}
    d['A']=string
    list(i).append(d)
list(i) = pd.DataFrame(list(i))
list(i)['Link'] = df['link'][i-1]
list(i)['Caption'] = df['caption'][i-1]
list(i)['Family_Type']=df['family_type'][i-1]
list(i)['Product_Type']=df['product_type'][i-1]
list(i).to_excel('list'+str(i).xlsx')

Variable Types:
print(type(dict1)) --> dict
print(type(dict1['a'])) --> list
print(type(string)) --> str
print(type(list1)) --> list

编辑尝试的代码：

d2 = {}
for i in range(1,100):
    d["list{0}".format(i)]=[]
        for i in range(1, 100):
            d["list{0}".format(i)] = comment.text
            d["list{0}".format(i).append(d2)
    d["list{0}".format(i)] = pd.DataFrame(d["list{0}".format(i)])
    d["list{0}".format(i)]['Link'] = df['link'][i-1]
    d["list{0}".format(i)]['Caption'] = df['caption'][i-1]
    d["list{0}".format(i)]['Family_Type'] = df['family_type'][i-1]
    d["list{0}".format(i)]['Product_Type'] = df['product_type'][i-1]
    d["list{0}".format(i)].to_excel('list'+str(i).xlsx')

返回错误

Answer 1

for i,_ in enumerate(df['link'], 1):
    listi = pd.DataFrame( [{'A':string} for a in locals()[f"dict{i}"]['a'] ] )
    listi['Link']=df['link'][i-1]
    listi['Caption']=df['caption'][i-1]
    listi['Family_Type']=df['family_type'][i-1]
    listi['Product_Type']=df['product_type'][i-1]
    listi.to_excel(f'list{i}.xlsx')

遍历字典和追加列表

1 个答案: