删除python中的特定行和相应文件

Question

我想删除90％的具有“转向”值qual的行为0.并且对于它们，中心，左侧和右侧都有相应的图像文件。我也想删除它们。 csv文件是这样的：

我编写了以下代码，至少得到转向值为0的文件。我只需要随机获取90％的文件并删除它们的代码。

with open('data/driving_log.csv') as csvfile:
    reader = csv.reader(csvfile)
    for i,line in enumerate(reader):
        lines.append(line)
        index.append(i)

lines = np.delete(lines,(0), axis = 0)
for i, line in enumerate(lines):
    #print(type(line[3].astype(np.float)))
    line_no.append(line[3].astype(np.float32))
    #print(line_no[i])
    if line_no[i]==0.0:
          # this gets the first column of the row.
        for j in range(3):
            source_path = line[j]
            filename = source_path.split('/')[-1]
            print(filename)
        count += 1

Answer 1

我认为这会做你想做的事情：

import csv
from random import randint
from os import remove

# Create a 2D list from which we can work with
lines = []
with open('data/driving_log.csv', newline='') as csvfile:
    reader = csv.reader(csvfile)
    for line in reader:
        lines.append(line)

# Find 10% of total lines (to keep), not including header row
numToKeep = round(sum(1 for i in lines if i[3] == '0') * 0.1)

# Save 10% of lines to a new 2D list
toKeep = []
for i in range(numToKeep):
    while True:
        index = randint(1, len(lines)-1)
        # Make sure we haven't already selected the same line
        if lines[index] not in toKeep and lines[index][3] == '0':
            toKeep.append(lines[index])
            break

# Deleting all files of the selected 90% of rows
for i, line in enumerate(lines):
    if i == 0:  # Omit the header row
        continue
    if lines[i][3] != '0':  # Keep rows that don't have a steering value of 0
        toKeep.append(lines[i])
    if line not in toKeep:
        print("Deleting: {}".format(line))
        for i in range(3):
            remove(line[i])

with open('data/driving_log.csv', 'w', newline='') as csvfile:
    writer = csv.writer(csvfile)
    writer.writerows([lines[0]])  # Put the header back in
    writer.writerows(toKeep)

我意识到这不是最优雅的解决方案。我不熟悉numpy，现在没有时间学习它，但这应该有效。