使用csv.DictReader：

import csv

with open('test.csv') as f:
    a = [{k: int(v) for k, v in row.items()}
        for row in csv.DictReader(f, skipinitialspace=True)]

将导致：

[{'col2': 2, 'col3': 3, 'col1': 1}, {'col2': 5, 'col3': 6, 'col1': 4}]

另一个更简单的答案：

import csv
with open("configure_column_mapping_logic.csv", "r") as f:
    reader = csv.DictReader(f)
    a = list(reader)
    print a

使用csv模块和列表解析：

import csv
with open('foo.csv') as f:
    reader = csv.reader(f, skipinitialspace=True)
    header = next(reader)
    a = [dict(zip(header, map(int, row))) for row in reader]
print a

输出：

[{'col3': 3, 'col2': 2, 'col1': 1}, {'col3': 6, 'col2': 5, 'col1': 4}]

很久之后才在这里回答，因为我没有看到任何更新/相关的答案。

df = pd.read_csv('Your csv file path')  
data = df.to_dict('records')
print( data )

# similar solution via namedtuple:    

import csv
from collections import namedtuple

with open('foo.csv') as f:
  fh = csv.reader(open(f, "rU"), delimiter=',', dialect=csv.excel_tab)
  headers = fh.next()
  Row = namedtuple('Row', headers)
  list_of_dicts = [Row._make(i)._asdict() for i in fh]

好吧，当其他人用聪明的方式来做的时候，我很天真地实现了它。我想我的方法的好处是不需要任何外部模块，尽管它可能会因为奇怪的值配置而失败。这里只是供参考：

a = []
with open("csv.txt") as myfile:
    firstline = True
    for line in myfile:
        if firstline:
            mykeys = "".join(line.split()).split(',')
            firstline = False
        else:
            values = "".join(line.split()).split(',')
            a.append({mykeys[n]:values[n] for n in range(0,len(mykeys))})

将CSV解析为字典列表的简单方法

with open('/home/mitul/Desktop/OPENEBS/test.csv', 'rb') as infile:
  header = infile.readline().split(",")
  for line in infile:
    fields = line.split(",")
    entry = {}
    for i,value in enumerate(fields):
      entry[header[i].strip()] = value.strip()
      data.append(entry)

我对你的问题的解决方案：

output = []

with open("test.csv", "r") as csv_file:
    data = csv_file.read().split("\n")

text = [data[i].split(",") for i in range(len(data))]

for i in range(1, len(text)) :
    output.append({text[0][num]:text[i][num] for num in range(len(text[i]))})

输出是dico的列表