我在与Python脚本相同的目录中有一个CSV文件,我想把这些数据转换成一个我以后可以使用的列表。我更喜欢使用Python的CSV模块。在阅读了模块的文档和有关它的问题之后,我仍然没有找到任何帮助。
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import csv
inputfile = 'inputfile.csv'
inputm = []
reader = csv.reader(inputfile)
for row in reader:
inputm.append(row)
point1,point2,point3,point4
0,1,4,1
0,1,1,1
0,1,4,1
0,1,0,0
0,1,2,1
0,0,0,0
0,0,0,0
0,1,3,1
0,1,4,1
它只返回我提供的文件名的字符串。
[['i'], ['n'], ['p'], ['u'], ['t'], ['f'], ['i'], ['l'], ['e']]
我希望将CSV文件的每一行作为子列表返回,而不是将文件名的每个字母作为子列表返回。
答案 0 :(得分:3)
您需要以读取模式打开文件,阅读内容! 也就是说,
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import csv
inputfile = 'inputfile.csv'
inputm = []
with open(inputfile, "rb") as f:
reader = csv.reader(f, delimiter="\t")
for row in reader:
inputm.append(row)
输出:
[['point1,point2,point3,point4'], ['0,1,4,1'], ['0,1,1,1'], ['0,1,4,1'], ['0,1,0,0'], ['0,1,2,1'], ['0,0,0,0'], ['0,0,0,0'], ['0,1,3,1'], ['0,1,4,1']]
答案 1 :(得分:1)
您确实需要open()
该文件:
inputfile = open('inputfile.csv')
您可能需要查看with
声明:
with open('inputfile.csv') as inputfile:
reader = csv.reader(inputfile)
inputm = list(reader)