您需要分割列表中的每个字符串：
import pandas as pd
df = pd.DataFrame([sub.split(",") for sub in l])
print(df)
输出：
0 1 2 3 4 5 6
0 AN 2__AS000 26 20150826113000 -283.000 20150826120000 -283.000
1 AN 2__A000 26 20150826113000 0.000 20150826120000 0.000
2 AN 2__AE000 26 20150826113000 -269.000 20150826120000 -269.000
3 AN 2__AE000 26 20150826113000 -255.000 20150826120000 -255.000
4 AN 2__AE00 26 20150826113000 -254.000 20150826120000 -254.000
如果您知道在csv中跳过多少行，则可以使用read_csv使用skiprows=lines_of_Metadata以下命令完成所有操作：
import pandas as pd
df = pd.read_csv("in.csv",skiprows=3,header=None)
print(df)
或者，如果元数据的每一行都以某个特定字符开头，则可以使用注释：
df = pd.read_csv("in.csv",header=None,comment="#")
如果你需要指定一个以上的字符，你可以组合itertools.takewhile这将下降开头为xxx：
import pandas as pd
from itertools import dropwhile
import csv
with open("in.csv") as f:
f = dropwhile(lambda x: x.startswith("#!!"), f)
r = csv.reader(f)
df = pd.DataFrame().from_records(r)
使用输入数据，添加以＃!!开头的几行：
#!! varIoUs
#!! Metadata
#!! lines
AN,2__AS000,26,20150826113000,-283.000,20150826120000,-283.000
AN,2__A000,26,20150826113000,0.000,20150826120000,0.000
AN,2__AE000,26,20150826113000,-269.000,20150826120000,-269.000
AN,2__AE000,26,20150826113000,-255.000,20150826120000,-255.000
AN,2__AE00,26,20150826113000,-254.000,20150826120000,-254.000
输出：
0 1 2 3 4 5 6
0 AN 2__AS000 26 20150826113000 -283.000 20150826120000 -283.000
1 AN 2__A000 26 20150826113000 0.000 20150826120000 0.000
2 AN 2__AE000 26 20150826113000 -269.000 20150826120000 -269.000
3 AN 2__AE000 26 20150826113000 -255.000 20150826120000 -255.000
4 AN 2__AE00 26 20150826113000 -254.000 20150826120000 -254.000
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