用Pythonic的方法可以讀取3個csv文件？

Question

def production(csvfile):
    # do something with csvfile
    print csvfile

production("somecsvfile.csv")

我想傳遞3個CSV文件文件，程序應為其輸出。當前我給出了一個CSV文件。

Answer 1

在函數中添加以下內容

fileList = csvfiles.split(",")
for files in filesList:
    data = read_csv(csvfile)
    ....
    ....


production("rswm20160901C.csv,rswm20160901E.csv,rswm20160901D.csv")

要么

import sys
def production(csvfile)
......
......

fileList = sys.argv[1:] # pass the functions in command line
for f in fileList:
    production(f)

Answer 2

我們不需要使情況復雜化。 應該這樣做：

def production(csvfile)
......
......

fileList = ['a.csv','b.csv','c.csv']
for f in fileList:
    production(f)

Answer 3

使用可以接受任意數量參數的python *args參數解包。 它適用於任何數量的參數

def production(*csvfile):
    for csv in csvfile:
        # do something with csvfile
        print(csv)

production("file1.csv", "file2.csv", "file3.csv")
# code above will printout 
# file1.csv 
# file2.csv 
# file3.csv
production("file1.csv")
# just print out 
# file1.csv

如果您希望它是精確的3個輸入文件，則可以定義3個輸入（如下面）或1個列表輸入，它們也適用於任何列表長度

def production(csvfile1, csvfile2, csvfile3):
    for csv in [csvfile1, csvfile2, csvfile3]:
        # do something with csvfile
        print(csv)