Question

试图从文本文件中读取十进制值会转换为16位二进制文件和二进制文件。

样本输入文件

代码：

def decimaltoBinary(filename,writefile):
    file = filename
    print(file)
    file_write = open(writefile,'wb')
    file_read = open(file, 'rb')
    for line in file_read:
        value = int(line)
        if value < 0:
            binary_value = bin((2**16) - abs(value))[2:].zfill(16)
            file_write.write(binary_value + "\n")
        else:
            binary_value = bin(int(value))[2:].zfill(16)
            file_write.write(binary_value + "\n")
    file_write.close()

decimaltoBinary(input_file.text,output_file.bin)

希望将转换后的十进制值写入二进制文件。非常感谢您的帮助

Answer 1

您可以使用the struct module：

data = [120,300,-250,13,-120] # you seem to have the reading part covered already
                              # using a list as data input for demo purposes
import struct

with open("f.bin","wb") as f: 
    for d in data:
        f.write(struct.pack('h', d)) # 2 byte integer aka short

with open("f.bin","rb") as f:
    print(f.read())  # b'x\x00,\x01\x06\xff\r\x00\x88\xff'

您只需要指定'h'即可进行简短（2字节整数）打包。

对于怪异的打印输出，怪怪的python-用较短的普通字符替换了“已知” \xXX代码-f.e. ',' => 0x2c或\r => \x0d等

如何转换从文本文件读取的整数并存储为具有16位整数的二进制文件？

1 个答案: