Question

我有一个csv这样的文件：

Time | Temperature
0.0  | 15.2
3.0  | 16.1
6.0  | 14.8
9.0  | 20.2

现在我想打印/返回它：(["Time", "Temperature"], [[0.0, 3.0, 6.0, 9.0], [15.2, 16.1, 14.8, 20.2]]) 但是python-docs的代码只返回了这个：

['Time', 'Temperature']
['0.0', '15.2']
['3.0', '16.1']
['6.0', '14.8']
['9.0', '20.2']

如何将其转换为一个列表？

编辑：我写了一个类，对于我的问题，我使用“parse_csv”方法：

import csv

class Table:
    def __init__(self, titles=None, cols=None, filename = None):
        self.titles = titles
        self.cols = cols
        self.filename = filename

    def __str__(self):
        return (str(self.titles) +', '+ str(self.cols))

    def __eq__(self, other):
        try:
            return (self.titles, self.cols) == (other.titles, other.cols)
        except AttributeError:
            return NotImplemented

    def parse_csv(self, filename):
        with open(filename, newline='') as csvfile:
            reader = csv.reader(csvfile)
            for row in reader:
                return row

EDIT2：我用这个检查代码：

titles = ["Time", "Temperature"]
filename1 = "simple_example.csv"
my_table1 = Table(titles, [[0.0, 3.0, 6.0, 9.0], [15.2, 16.1, 14.8, 20.2]])
my_table2 = Table(filename = "simple_example.csv")
assert my_table2 == my_table3

Answer 1

我假设您要将标题[column1和column2]作为元组返回。你可以用熊猫轻松完成。

import pandas as pd
def parse_csv(file_name):
    df = pd.read_csv(file_name)
    time = df['Time'].as_matrix().tolist() #returns the time column as a list
    temperature = df['Temperature'].as_matrix().tolist() #returns the temperature column as a list
    header = list(df.columns.values) #returns the header values as a list
    return (header, [time, temperature]) #returning a tuple

Answer 2

这是你应该如何定义你的类并初始化它。

import csv

class Table:
    def __init__(self, titles=None, cols=None, filename = None):
        self.titles = titles
        self.cols = cols
        self.filename = filename

    def __str__(self):
        return (str(self.titles) +', '+ str(self.cols))

    def __eq__(self, other):
        try:
            return (self.titles, self.cols) == (other.titles, other.cols)
        except AttributeError:
            return NotImplemented

    def parse_csv(self, filename=None):
        if not filename:
            filename=self.filename
        with open(filename, newline='') as csvfile:
            reader = csv.reader(csvfile, delimiter='|')
            line = [[a.strip() for a in line] for line in reader]
            self.titles= line[0]
            self.cols = [list(a) for a in zip(*line[1:])]
            self.cols = [[float(b) for b in c] for c in self.cols]

titles = ["Time", "Temperature"]
filename1 = "simple_example.csv"
my_table1 = Table(titles, cols=[[0.0, 3.0, 6.0, 9.0], [15.2, 16.1, 14.8, 20.2]])
my_table2 = Table(filename="simple_example.csv")
my_table2.parse_csv()
print my_table1 == my_table2

输出：

True

Answer 3

我想这就是你想要实现的目标：

import csv

def parse_csv(filename):
    with open(filename) as f:
        reader = csv.reader(f)
        header = next(reader)
        cols = list(zip(*reader))
    return [header, cols]

next返回reader迭代器的第一个元素（标题行），然后zip转换列向量中的剩余行。

Answer 4

使用zip()函数的解决方案：

with open('test.csv', 'r') as fh:
    contents = list(csv.reader(fh, delimiter='|', quotechar='|'))
    result = []
    result.append(contents.pop(0))
    result.append([list(s) for s in list(zip(*contents))])

print(result)

输出：

[['Time ', ' Temperature'], [['0.0  ', '3.0  ', '6.0  ', '9.0  '], [' 15.2', ' 16.1', ' 14.8', ' 20.2']]]

Answer 5

a=open('filepath','r')
final,timel,templ,count=[],[],[],0

for l in a:
    c=l.split(',')
    if count >=1: 
        timel.append(float(c[0]))
        templ.append(float(c[1]))
    if count==0:        #this is just to append the header in the final result
        c[1]=c[1][:-1]  #to remove the '\n'
        final.append(c)
        count+=1  

final.append([timel,templ])
final=tuple(final)

print(final)

Answer 6

import numpy as np
y = np.array(x)
z = [y[0],y[1:,0],y[1:,1]]
res = [i.tolist() for i in z]
print res

只返回csv的列

6 个答案: