创建 TXT 文件并在 Python 中寻求职位
Creating a TXT file and seeking a position in Python
我已经给出了以下变量:
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
我想创建一个 .TXT 文件,它看起来像这样,带有制表符分隔值:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
500 1000 5 8 9 0
1000 1500 6 7 11 19
1500 2000 1 4 5 12
2000 2500 -5 8 9 0
2500 3000 -6 7 11 19
3000 3500 1 4 5 12
3500 4000 -5 8 9 0
4000 4500 -6 7 11 19
我写了下面的代码:
#!/usr/bin/env python3
import os
from datetime import datetime
import time
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
filename = f"{datetime.now().strftime('%Y%m%d_%H%M%S')}_{signal1}_results.TXT"
with open(filename, 'w') as f:
# write the bin1 range
f.write('\n\n\n')
f.write('\t\t\t\t')
f.write(signal1 + '>=')
for bin in bins1[:-1]:
f.write('\t' + str(bin))
f.write('\n')
f.write('\t\t\t\t')
f.write(signal1 + '<')
for bin in bins1[1:]:
f.write('\t' + str(bin))
f.write('\n')
# write the bin2 range
f.write('\t\t')
f.write(signal2 + '>=' + '\t' + signal2 + '<' + '\n')
f.write('\t\t')
# store the cursor position from where hist result will be written line by line
track_cursor_pos = []
curr = bins2[0]
for next in bins2[1:]:
f.write(str(curr) + '\t' + str(next))
track_cursor_pos.append(f.tell())
f.write('\n\t\t')
curr = next
f.write('\n')
print(track_cursor_pos)
i = 0
# Everything is fine until here
# Code below doesn't work as expected!?
for result in hist_result:
f.seek(track_cursor_pos[i], os.SEEK_SET)
for r in result:
f.write('\t' + str(r))
f.write('\n')
i += 1
但是,这是生成内容如下所示的 TXT 文件:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
0 -5 8 9 0
00 -6 7 11 19
1 4 5 12
00 -5 8 9 0
00 -6 7 11 19
1 4 5 12
00 -5 8 9 0
00 -6 7 11 19
我想我没有正确使用 f.seek()
。任何建议将不胜感激。提前致谢。
您无需在文件内部查找即可打印您的数据:
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
with open('data.txt', 'w') as f_out:
print('\t{signal1}>=\t{bins}'.format(signal1=signal1, bins='\t'.join(map(str,bins1[:-1]))), file=f_out)
print('\t{signal1}<\t{bins}'.format(signal1=signal1, bins='\t'.join(map(str,bins1[1:]))), file=f_out)
print('{signal2}>=\t{signal2}<'.format(signal2=signal2))
for a, b, data in zip(bins2[:-1], bins2[1:], hist_result):
print(a, b, *data, sep='\t', file=f_out)
创建 data.txt
:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
500 1000 -5 8 9 0
1000 1500 -6 7 11 19
1500 2000 1 4 5 12
2000 2500 -5 8 9 0
2500 3000 -6 7 11 19
3000 3500 1 4 5 12
3500 4000 -5 8 9 0
4000 4500 -6 7 11 19
我已经给出了以下变量:
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
我想创建一个 .TXT 文件,它看起来像这样,带有制表符分隔值:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
500 1000 5 8 9 0
1000 1500 6 7 11 19
1500 2000 1 4 5 12
2000 2500 -5 8 9 0
2500 3000 -6 7 11 19
3000 3500 1 4 5 12
3500 4000 -5 8 9 0
4000 4500 -6 7 11 19
我写了下面的代码:
#!/usr/bin/env python3
import os
from datetime import datetime
import time
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
filename = f"{datetime.now().strftime('%Y%m%d_%H%M%S')}_{signal1}_results.TXT"
with open(filename, 'w') as f:
# write the bin1 range
f.write('\n\n\n')
f.write('\t\t\t\t')
f.write(signal1 + '>=')
for bin in bins1[:-1]:
f.write('\t' + str(bin))
f.write('\n')
f.write('\t\t\t\t')
f.write(signal1 + '<')
for bin in bins1[1:]:
f.write('\t' + str(bin))
f.write('\n')
# write the bin2 range
f.write('\t\t')
f.write(signal2 + '>=' + '\t' + signal2 + '<' + '\n')
f.write('\t\t')
# store the cursor position from where hist result will be written line by line
track_cursor_pos = []
curr = bins2[0]
for next in bins2[1:]:
f.write(str(curr) + '\t' + str(next))
track_cursor_pos.append(f.tell())
f.write('\n\t\t')
curr = next
f.write('\n')
print(track_cursor_pos)
i = 0
# Everything is fine until here
# Code below doesn't work as expected!?
for result in hist_result:
f.seek(track_cursor_pos[i], os.SEEK_SET)
for r in result:
f.write('\t' + str(r))
f.write('\n')
i += 1
但是,这是生成内容如下所示的 TXT 文件:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
0 -5 8 9 0
00 -6 7 11 19
1 4 5 12
00 -5 8 9 0
00 -6 7 11 19
1 4 5 12
00 -5 8 9 0
00 -6 7 11 19
我想我没有正确使用 f.seek()
。任何建议将不胜感激。提前致谢。
您无需在文件内部查找即可打印您的数据:
signal1 = 'speed'
bins1 = [0, 10, 20, 30, 40]
signal2 = 'rpm'
bins2 = [0, 500, 1000, 1500, 2000, 2500, 3000, 3500, 4000, 4500]
hist_result = [ [1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
[1, 4, 5, 12],
[-5, 8, 9, 0],
[-6, 7, 11, 19],
]
with open('data.txt', 'w') as f_out:
print('\t{signal1}>=\t{bins}'.format(signal1=signal1, bins='\t'.join(map(str,bins1[:-1]))), file=f_out)
print('\t{signal1}<\t{bins}'.format(signal1=signal1, bins='\t'.join(map(str,bins1[1:]))), file=f_out)
print('{signal2}>=\t{signal2}<'.format(signal2=signal2))
for a, b, data in zip(bins2[:-1], bins2[1:], hist_result):
print(a, b, *data, sep='\t', file=f_out)
创建 data.txt
:
speed>= 0 10 20 30
speed< 10 20 30 40
rpm>= rpm<
0 500 1 4 5 12
500 1000 -5 8 9 0
1000 1500 -6 7 11 19
1500 2000 1 4 5 12
2000 2500 -5 8 9 0
2500 3000 -6 7 11 19
3000 3500 1 4 5 12
3500 4000 -5 8 9 0
4000 4500 -6 7 11 19