I have a question about how to save a list as a CSV file in my way:
I have a large list with many columns:
ex. ['2020-03-01T21:46:56Z', '059d422b-9da1-4be4-836b-dca71e44ff24', '23.595858333333336', '37.95288666666667', '', '', '0', '80', 'PERIOXH', 0.0005636937949292874]
Although I don't want to save the last column : 0.0005636937949292874
How can I do it? Here is my code:
with open(save_file, 'w', newline='\n') as f:
writer = csv.writer(f)
writer.writerows(result_buffer)
you can iterate over your list and remove the last item for each line:
with open(save_file, 'w', newline='\n') as lines:
writer = csv.writer(f)
for line in lines:
writer.writerow(line[:-1])
You can slice your list
list = ['2020-03-01T21:46:56Z', '059d422b-9da1-4be4-836b-dca71e44ff24', '23.595858333333336', '37.95288666666667', '', '', '0', '80', 'PERIOXH', 0.0005636937949292874]
print(list[0:-1])
>>>['2020-03-01T21:46:56Z', '059d422b-9da1-4be4-836b-dca71e44ff24', '23.595858333333336', '37.95288666666667', '', '', '0', '80', 'PERIOXH']
I used slicing.
my_list = ['2020-03-01T21:46:56Z', '059d422b-9da1-4be4-836b-dca71e44ff24',
'23.595858333333336', '37.95288666666667', '', '', '0', '80', 'PERIOXH',
0.0005636937949292874]
import csv
with open('result.csv', 'w', newline='') as f:
writer = csv.writer(f)
writer.writerow(my_list[:-1])
Output:
2020-03-01T21:46:56Z,059d422b-9da1-4be4-836b-dca71e44ff24,23.595858333333336,37.95288666666667,,,0,80,PERIOXH
