加载腌制文件时出错

Question

我在 python3 中加载腌制文件时遇到问题。见以下代码：

#!/usr/bin/env python3

import csv,operator
import pickle
import os.path

pathToBin = "foo/bar/foobar.bin"
pathToCSV = "foo/bar/foobar.csv" 

if os.path.isfile(pathToBin):
        print("Binary file already on harddrive, loading from there")
        transactions = pickle.loads( open( pathToBin, "rb" ))
else:
        csvIn = open(pathToCSV,'r')
        reader = csv.reader(csvIn)
        header = next(reader)
        header = header[0].split(";")

        print("Reading file")
        transactions = []
        for row in reader:
                # read file. 
        # transactions contains now lists of strings: transactions = [ ["a","b","c"], ["a2","b2","c3"], ...]
        print("Dumping python file to harddrive")
        myPickleFile = open(pathToBin,'wb')
        pickle.dump(transactions, myPickleFile, protocol=pickle.HIGHEST_PROTOCOL)

# do some more computation

保存文件没有任何问题。但是，加载它会给我以下错误：

transactions = pickle.loads( open( pathToBin, "rb" ))
TypeError: '_io.BufferedReader' does not support the buffer interface

自从我使用 python3 以来，字符串的处理方式有所不同。因此，我专门为 saving/loading 提供了 "b" 选项。有人知道为什么这行不通吗？

Answer 1

您想使用 load:

transactions = pickle.load(open( pathToBin, "rb" ))

从您打开的文件句柄中读取。 loads 接受 bytes 而不是文件句柄（意味着：加载 "string"，现在在 python 3 升级 string/bytes 处理后加载 "bytes"：

transactions = pickle.loads(open( pathToBin, "rb" ).read())

读取文件返回的字节。

在这种情况下，我会为您推荐第一个选项。第二个选项适用于更复杂的情况。

旁白：最好使用 with 上下文来控制文件何时关闭

with open( pathToBin,"rb") as f:
    transactions = pickle.load(f)

加载腌制文件时出错

Error loading pickled file

pickle

python-3.x