解決你的問題是下面的find_filesets()
方法。我也包含了一個基於MaxNoe's answer的CSV合併方法。
#!/usr/bin/env python
import glob
import random
import os
import pandas
def rm_minus_rf(dirname):
for r,d,f in os.walk(dirname):
for files in f:
os.remove(os.path.join(r, files))
os.removedirs(r)
def create_testfiles(path):
rm_minus_rf(path)
os.mkdir(path)
random.seed()
for i in range(10):
n = random.randint(10000,99999)
for j in range(random.randint(0,20)):
# year may repeat, doesn't matter
year = 2015 - random.randint(0,20)
with open("{}/{}-{}.csv".format(path, n, year), "w"):
pass
def find_filesets(path="."):
csv_files = {}
for name in glob.glob("{}/*-*.csv".format(path)):
# there's almost certainly a better way to do this
key = os.path.splitext(os.path.basename(name))[0].split('-')[0]
csv_files.setdefault(key, []).append(name)
for key,filelist in csv_files.items():
print key, filelist
# do something with filelist
create_merged_csv(key, filelist)
def create_merged_csv(key, filelist):
with open('{}-aggregate.csv'.format(key), 'w+b') as outfile:
for filename in filelist:
df = pandas.read_csv(filename, header=False)
df.to_csv(outfile, index=False, header=False)
TEST_DIR_NAME="testfiles"
create_testfiles(TEST_DIR_NAME)
find_filesets(TEST_DIR_NAME)
@ lego-stormtroopr:我的答案是完整的,只有47行。另外,儘管各向異性可以更好地解決問題,但他非常清楚自己想要完成的任務:'給定一個包含以Prefix-Year.csv模式命名的CSV文件的目錄,創建一組名爲Prefix-aggregate的CSV文件。 csv其中每個聚合文件是具有相同前綴的所有CSV文件的組合。 – Harvey
@mariano:管理員是否會自動收到有關他們標記/暫停/等問題的評論的通知? – Harvey
@ rajesh-jadav:上面的「幫助中心」鏈接表示如果原始海報進行編輯,問題將被放入審閱隊列中。 – Harvey