你需要拆分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
如果你知道在csv文件中需要跳过多少行元数据,你可以使用skiprows=lines_of_metadata
参数在read_csv函数中完成此操作:
import pandas as pd
df = pd.read_csv("in.csv",skiprows=3,header=None)
print(df)
或者,如果元数据每行都以特定字符开头,您可以使用注释:
df = pd.read_csv("in.csv",header=None,comment="#")
如果您需要指定多个字符,则可以结合使用 itertools.takewhile
,它将删除以 xxx
开头的行:
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)
使用您的输入数据添加一些以 #!! 开头的行:
#!! 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
print(mylist[:3])
的输出。 - EdChum