我的文本文件包含以下行:
Cycle 0 DUT 2 Bad Block : 2,4,6,7,8,10,12,14,16,18,20,22,24,26,28
Cycle 0 DUT 3 Bad Block : 4,6,8,10,12,14,16,18,20,22,24,26
Cycle 0 DUT 4 Bad Block : 4,6,8,10,12,14,16,18,20,22,24,26
Cycle 1 DUT 2 Bad Block : 2,4,6,7,8,10,12,14,16,18,20,22,24,26,28
Cycle 1 DUT 3 Bad Block : 4,6,8,10,12,14,16,18,20,22,24,26,28,30,32
我想将Cycle 0 DUT 2
文本行(用逗号分隔冒号后的数字)与Cycle 1 DUT 2
文本行(冒号用逗号分隔后的数字)进行比较并得到差异,然后比较{{1将文本行添加到Cycle 0 DUT 3
文本行并获取差异或唯一值。
答案 0 :(得分:1)
我想你想把关键字改为DUT
数字:
import re
dut_data = {}
cycle_dut = re.compile('^Cycle\s+(\d)\s+DUT\s+(\d)\s+Bad Block\s*:\s*(.*)$')
with open(inputfile, 'r') as infile:
for line in infile:
match = cycle_dut.search(line)
if match:
cycle, dut, data = match.groups()
data = [int(v) for v in data.split(',')]
if cycle == '0':
# Store cycle 0 DUT values keyed on the DUT number
dut_data[dut] = data
else:
# Compare against cycle 0 data, if the same DUT number was present
cycle_0_data = dut_data.get(dut)
if cycle_0_data is not None:
# compare cycle_0_data and data here
print 'DUT {} differences: {}'.format(dut, ','.join([str(v) for v in sorted(set(cycle_0_data).symmetric_difference(data))]))
我使用快速设置差异来打印差异,这可能需要精炼。
对于您的样本数据,打印:
DUT 2 differences:
DUT 3 differences: 28,30,32