使用CSV文件将文件移动到不同的目录

Question

我有一个机器学习数据集中的大型1000+文件目录，但是这些文件有不同的质量(玫瑰图片与守护花图片，以保持简单)。我有这个CSV文件，其中包含数据集中每一项的文件名以及它们的分类(玫瑰对雏菊)。我如何读取这个CSV文件，并告诉我的文件管理器将所有玫瑰照片移至一个目录，将所有雏菊照片移至另一个目录？我是否需要使用Bash脚本，或者这是已经构建在Nautilus中的东西？

Answer 1

好吧，我和一个朋友用Python写了一个脚本，很好地解决了这个问题。

# Import csv
import csv
# Import os
import os

# Main Function
def main():
# Open dataset file
dataset = open('dataset.csv', newline='')

# Initialize csvreader for dataset
reader = csv.reader(dataset)

# Read data from reader
data = list(reader)

# Variables for progress counter
lines = len(data)
i = 0

# Analyze data in dataset
for row in data:
    # Assign image name and state to variables
    image = row[0] + '.jpeg'
    state = row[1]

    # Print image information
    print('({}/{}) Processing image ({}): {}'.format(i + 1, lines, state, image))

    # Increment i
    i += 1

    # Determine action to perform
    if state is '0':
        # Attempt to move the file
        try:
            # Move the file to nosymptoms/
            os.rename(image, 'nosymptoms/' + image)
            # Inform the user of action being taken
            print(' -> Moved to nosymptoms/')
        except FileNotFoundError:
            # Inform the user of the failure
            print(' -> Failed to find file')
    elif state in ['1', '2', '3', '4']:
        # Attempt to move the file
        try:
            # Move the file to nosymptoms/
            os.rename(image, 'symptoms/' + image)
            # Inform the user of action being taken
            print(' -> Moved to symptoms/')
        except FileNotFoundError:
            # Inform the user of the failure
            print(' -> Failed to find file')

# Execute main function if name is equal to main
if __name__ == '__main__':
main()

这往往更好，因为我有更多的类别来处理with...hopefully，这将适用于任何有同样问题的人。