Question

每个文件夹都有一个csv用于一年中的每个月（1.csv，2.csv，3.csv等），脚本会创建一个数据框，将所有12个csv的第9列合并到xlsx表中名为concentrated.xlsx。它可以工作，但一次只能用于一个目录

files = glob['2014/*.csv']
sorted_files = natsorted(files)

def read_9th(fn):
    return pd.read_csv(fn, usecols=[9], names=headers)

big_df = pd.concat([read_9th(fn) for fn in sorted_files], axis=1)
writer = pd.ExcelWriter('concentrated.xlsx', engine='openpyxl') 
big_df.to_excel(writer,'2014')

writer.save()

是否可以自动为每个目录创建数据框，而无需为每个文件夹手动创建一个数据框，如下所示：

files14 = glob['2014/*.csv']
files15 = glob['2015/*.csv']

sorted_files14 = natsorted(files14)
sorted_files15 = natsorted(files15)

def read_9th(fn):
    return pd.read_csv(fn, usecols=[9], names=headers)

big_df = pd.concat([read_9th(fn) for fn in sorted_files14], axis=1)
big_df1 = pd.concat([read_9th(fn) for fn in sorted_files15], axis=1)
writer = pd.ExcelWriter('concentrated.xlsx', engine='openpyxl') 
big_df.to_excel(writer,'2014')
big_df1.to_excel(writer,'2015')

writer.save()

Answer 1

如果您获得了要处理的文件夹列表，例如

folders = os.listdir('.')
# or 
folders = ['2014', '2015', '2016']

您可以执行以下操作：

writer = pd.ExcelWriter('concentrated.xlsx', engine='openpyxl')
for folder in folders:
    files = glob('%s/*.csv' % folder)
    sorted_files = natsorted(files)

    big_df = pd.concat([read_9th(fn) for fn in sorted_files], axis=1) 
    big_df.to_excel(writer, folder)

writer.save()

自动为每个文件夹创建数据框

1 个答案: