讀2的從二進制文件的補充（32位int）

Question

好吧，首先，大家好:) 我運行的Python 3.2的x86 @窗口

所以，我已經有了一個二進制文件（6 412字節）具有如下結構：

header: 256 bytes 
maxZ: 4 bytes (1 32-bit word) 
Zvalues: 2048 bytes (512 32-bit words) 
maxX: *same as maxZ* 
Xvalues: *same as Zvalues* 
maxY: *same as maxZ* 
Yvalues: *same as Zvalues* 

我知道（由於格式規範），該號碼存儲爲32位整數寫成2的補

說我不需要讀取頭，這就是我所做的（我有1539個4字節的樣本可供讀取）：

import struct 
file = open('C:/02.adb', 'rb') 
file.seek(256) 
fmt = '<i' 
nbytes = 4 
for i in range(1539): 
    packed_data = file.read(nbytes) 
    unpacked_data = struct.unpack(fmt, packed_data) 
    print('Sample: ',i,'Value: ',unpacked_data[0]) 
file.close() 

讀取文件並將其內容打印到標準輸出中，但標記有問題。我想我不是在閱讀數字作爲2的補充在這裏...我如何改進腳本？

TNX的關注

另外（12年10月23日）：

我意識到，通過編寫一個更大的外部腳本的短版我剛剛解決的主要問題本身 - 我自己mindlessness。

可以從文件頭中讀取的樣本數並不是確切的存儲樣本總數，而是排除存儲在每個通道數據的第一個字中的最大幅度的ZXY值組合數塊。

這裏的腳本的最後短版，它的輸出（是的，簽署的int蟒蛇表示是「2的補」）：

import struct 
file = open('C:/02.adb', 'rb') 
file.seek(256) 
transfer_ratio = 31446.540881 
fmt = '<i' 
nbytes = 4 
samples = [] 
for i in range(1539): 
    packed_data = file.read(nbytes) 
    unpacked_data = struct.unpack(fmt, packed_data) 
    samples.append(int(unpacked_data[0])/transfer_ratio) 
for i in range(32): 
    print('Ch1_val: ',samples[i+1],' Ch2_val: ',samples[i+1+513],' Ch3_val: ',samples[i+1+(513*2)]) 
file.close() 

輸出：

Ch1_val: 42.20127119933322 Ch2_val: 36.47237399942374 Ch3_val: -3.4996535999447054 
Ch1_val: 41.37017819934635 Ch2_val: 36.64441199942102 Ch3_val: -4.35822179993114 
Ch1_val: 42.43585979932951 Ch2_val: 34.9874729994472 Ch3_val: -4.125827399934812 
Ch1_val: 44.08278179930349 Ch2_val: 34.42912859945602 Ch3_val: -4.1604257999342655 
Ch1_val: 44.50289159929685 Ch2_val: 35.507053199438985 Ch3_val: -3.092836199951133 
Ch1_val: 43.69110119930968 Ch2_val: 35.53335179943857 Ch3_val: -2.613991799958699 
Ch1_val: 42.73662419932476 Ch2_val: 34.1708807994601 Ch3_val: -5.254886399916972 
Ch1_val: 42.14097839933417 Ch2_val: 34.08460739946146 Ch3_val: -8.582724599864394 
Ch1_val: 41.982105599336684 Ch2_val: 35.58544019943775 Ch3_val: -8.84688719986022 
Ch1_val: 42.503816399328436 Ch2_val: 36.06170879943022 Ch3_val: -6.777056999892922 
Ch1_val: 43.17568679931782 Ch2_val: 35.291926199442386 Ch3_val: -4.984618199921243 
Ch1_val: 42.4303583993296 Ch2_val: 36.03750899943061 Ch3_val: -4.502275799928864 
Ch1_val: 39.92264219936922 Ch2_val: 38.354075399394006 Ch3_val: -6.050744999904398 
Ch1_val: 37.65501599940505 Ch2_val: 38.48181599939198 Ch3_val: -9.010943399857627 
Ch1_val: 37.04642759941466 Ch2_val: 35.152642199444585 Ch3_val: -9.89107199984372 
Ch1_val: 36.571876199422164 Ch2_val: 32.21988719949093 Ch3_val: -7.850465999875962 
Ch1_val: 34.488403799455085 Ch2_val: 32.62164839948458 Ch3_val: -7.712485799878142 
Ch1_val: 32.00259779949436 Ch2_val: 34.05570119946192 Ch3_val: -11.215064999822802 
Ch1_val: 31.132422599508107 Ch2_val: 33.44755799947153 Ch3_val: -12.461179799803112 
Ch1_val: 31.245853199506314 Ch2_val: 31.730771399498654 Ch3_val: -8.723057999862176 
Ch1_val: 30.40820939951955 Ch2_val: 31.03355639950967 Ch3_val: -6.433076399898357 
Ch1_val: 28.474165199550107 Ch2_val: 30.714824999514704 Ch3_val: -9.730672799846255 
Ch1_val: 26.481449999581592 Ch2_val: 29.53720739953331 Ch3_val: -13.227496199791005 
Ch1_val: 24.79611359960822 Ch2_val: 28.60448159954805 Ch3_val: -12.1075955998087 
Ch1_val: 23.74938479962476 Ch2_val: 28.95253259954255 Ch3_val: -9.411305399851301 
Ch1_val: 23.92663799962196 Ch2_val: 29.24305739953796 Ch3_val: -8.856490799860067 
Ch1_val: 24.37285559961491 Ch2_val: 28.72398599954616 Ch3_val: -9.611549999848137 
Ch1_val: 23.155774199634138 Ch2_val: 29.22521759953824 Ch3_val: -10.701304199830918 
Ch1_val: 20.822671799671003 Ch2_val: 31.46641799950283 Ch3_val: -12.57088979980138 
Ch1_val: 20.15302739968158 Ch2_val: 32.95071479947938 Ch3_val: -13.581175799785417 
Ch1_val: 21.440195999661245 Ch2_val: 32.28628559948987 Ch3_val: -11.66643419981567 
Ch1_val: 21.51502139966006 Ch2_val: 31.7910959994977 Ch3_val: -8.843802599860268 

Answer 1

我相當肯定Python使用two's complement numbers作爲有符號整數。一個常見的問題是endianness。規範應該提及數字是小端還是大端。在Python中，您可以同時加載<i和>i。