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

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

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)
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)

基于希德的好答案。

识别列缺失或未对齐的问题

在连接之前,您可以将CSV文件加载到一个中间字典中,该字典根据文件名(以dict_of_df['filename.csv']的形式)访问每个数据集。这样的字典可以帮助您识别异构数据格式的问题,例如当列名没有对齐时。

导入模块并定位文件路径:

import os
import glob
import pandas
from collections import OrderedDict
path =r'C:\DRO\DCL_rawdata_files'
filenames = glob.glob(path + "/*.csv")

注意:OrderedDict不是必需的,但它将保持文件的顺序,这可能对分析有用。

加载CSV文件到字典中。然后连接:

dict_of_df = OrderedDict((f, pandas.read_csv(f)) for f in filenames)
pandas.concat(dict_of_df, sort=True)

键为文件名称f,值为CSV文件的数据帧内容。

除了使用f作为字典键,你还可以使用os.path.basename(f)或其他os.path.basename(f)。方法将字典中键的大小减少到仅相关的较小部分。

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

考虑使用convtools库,它提供了大量数据处理原语,并在底层生成简单的临时代码。 它不应该比熊猫/极地快,但有时它可以。

例如,你可以连接到一个CSV文件进一步重用-这是代码:

import glob

from convtools import conversion as c
from convtools.contrib.tables import Table
import pandas as pd


def test_pandas():
    df = pd.concat(
        (
            pd.read_csv(filename, index_col=None, header=0)
            for filename in glob.glob("tmp/*.csv")
        ),
        axis=0,
        ignore_index=True,
    )
    df.to_csv("out.csv", index=False)
# took 20.9 s


def test_convtools():
    table = None
    for filename in glob.glob("tmp/*.csv"):
        table_ = Table.from_csv(filename, header=False)
        if table is None:
            table = table_
        else:
            table = table.chain(table_)

    table.into_csv("out_convtools.csv", include_header=False)
# took 15.8 s

当然,如果你只是想获得一个数据帧而不写入一个连接文件,它将相应地花费4.63秒和10.9秒(pandas在这里更快,因为它不需要压缩列来写入回)。