你可以用这个，它很酷:

import dataconverters.commas as commas
filename = 'test.csv'
with open(filename) as f:
      records, metadata = commas.parse(f)
      for row in records:
            print 'this is row in dictionary:'+rowenter code here

import csv
reader = csv.reader(open('filename.csv', 'r'))
d = {}
for row in reader:
   k, v = row
   d[k] = v

假设你有一个这样结构的CSV:

"a","b"
1,2
3,4
5,6

你希望输出是:

[{'a': '1', ' "b"': '2'}, {'a': '3', ' "b"': '4'}, {'a': '5', ' "b"': '6'}]

zip函数(还没有提到)非常简单，而且非常有用。

def read_csv(filename):
    with open(filename) as f:
        file_data=csv.reader(f)
        headers=next(file_data)
        return [dict(zip(headers,i)) for i in file_data]

如果你更喜欢熊猫，它也可以很好地做到这一点:

import pandas as pd
def read_csv(filename):
    return pd.read_csv(filename).to_dict('records')

我相信您正在寻找的语法如下:

import csv

with open('coors.csv', mode='r') as infile:
    reader = csv.reader(infile)
    with open('coors_new.csv', mode='w') as outfile:
        writer = csv.writer(outfile)
        mydict = {rows[0]:rows[1] for rows in reader}

另外，对于python <= 2.7.1，你需要:

mydict = dict((rows[0],rows[1]) for rows in reader)

如果你可以使用numpy包，那么你可以像下面这样做:

import numpy as np

lines = np.genfromtxt("coors.csv", delimiter=",", dtype=None)
my_dict = dict()
for i in range(len(lines)):
   my_dict[lines[i][0]] = lines[i][1]

如果你有:

csv中只有一个键和一个作为键的值 不想导入其他包 想要一次创造字典

这样做:

mydict = {y[0]: y[1] for y in [x.split(",") for x in open('file.csv').read().split('\n') if x]}

它能做什么?

它使用列表推导式来分割行，最后一个“if x”用于忽略空行(通常在末尾)，然后使用字典推导式将空行解压缩到字典中。