3 回答
![?](http://img1.sycdn.imooc.com/5333a0780001a6e702200220-100-100.jpg)
TA贡献1829条经验 获得超13个赞
假设 n=3 并使用列表理解:
import pandas as pd
filelist = ['file1','file2','file3']
filelist1=[f for f in filelist for i in range(3)]
df1 = pd.DataFrame(filelist1, columns=['labels'])
print(df1)
输出:
labels
0 file1
1 file1
2 file1
3 file2
4 file2
5 file2
6 file3
7 file3
8 file3
![?](http://img1.sycdn.imooc.com/533e4d470001a00a02000200-100-100.jpg)
TA贡献1815条经验 获得超6个赞
使用熊猫如下:
import pandas as pd
filelist = ['file1','file2','file3']
df = pd.DataFrame({"labels":filelist})
df
输出如下:
labels
0 file1
1 file2
2 file3
![?](http://img1.sycdn.imooc.com/545869390001b87802200220-100-100.jpg)
TA贡献2041条经验 获得超4个赞
您可以定义一个函数来执行此操作
def createDF(input_list, n_iteration):
data = sorted(input_list * n_iteration)
df = pd.DataFrame(data={'Labels': data})
return df
createDF(filelist, 3)
输出
Labels
0 file1
1 file1
2 file1
3 file2
4 file2
5 file2
6 file3
7 file3
8 file3
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