我想从目录中读取几个CSV文件到熊猫，并将它们连接到一个大的DataFrame。不过我还没弄明白。以下是我目前所掌握的:
import glob
import pandas as pd
# Get data file names
path = r'C:\DRO\DCL_rawdata_files'
filenames = glob.glob(path + "/*.csv")
dfs = []
for filename in filenames:
dfs.append(pd.read_csv(filename))
# Concatenate all data into one DataFrame
big_frame = pd.concat(dfs, ignore_index=True)
我想我在for循环中需要一些帮助?
当前回答
import pandas as pd
import glob
path = r'C:\DRO\DCL_rawdata_files' # use your path
file_path_list = glob.glob(path + "/*.csv")
file_iter = iter(file_path_list)
list_df_csv = []
list_df_csv.append(pd.read_csv(next(file_iter)))
for file in file_iter:
lsit_df_csv.append(pd.read_csv(file, header=0))
df = pd.concat(lsit_df_csv, ignore_index=True)
其他回答
如果多个CSV文件被压缩，您可以使用zipfile读取所有文件并按以下方式连接:
import zipfile
import pandas as pd
ziptrain = zipfile.ZipFile('yourpath/yourfile.zip')
train = []
train = [ pd.read_csv(ziptrain.open(f)) for f in ziptrain.namelist() ]
df = pd.concat(train)
如果出现未命名列的问题，请使用此代码沿x轴合并多个CSV文件。
import glob
import os
import pandas as pd
merged_df = pd.concat([pd.read_csv(csv_file, index_col=0, header=0) for csv_file in glob.glob(
os.path.join("data/", "*.csv"))], axis=0, ignore_index=True)
merged_df.to_csv("merged.csv")
另一个带有列表理解的一行程序，允许使用read_csv参数。
df = pd.concat([pd.read_csv(f'dir/{f}') for f in os.listdir('dir') if f.endswith('.csv')])
darindaCoder的答案的替代方案:
path = r'C:\DRO\DCL_rawdata_files' # use your path
all_files = glob.glob(os.path.join(path, "*.csv")) # advisable to use os.path.join as this makes concatenation OS independent
df_from_each_file = (pd.read_csv(f) for f in all_files)
concatenated_df = pd.concat(df_from_each_file, ignore_index=True)
# doesn't create a list, nor does it append to one
使用map的一行代码，但是如果你想指定额外的参数，你可以这样做:
import pandas as pd
import glob
import functools
df = pd.concat(map(functools.partial(pd.read_csv, sep='|', compression=None),
glob.glob("data/*.csv")))
注意:map本身不允许您提供额外的参数。
