使用python剥离CSV中的零

Question

你好，我有一个csv文件，我需要用python删除零:

python中的第6列，第5列默认为7位。这个

AFI12001,01,C-,201405,P,0000430,2,0.02125000,US,60.0000
AFI12001,01,S-,201404,C,0001550,2,0.03500000,US,30.0000

我需要删除前面的0然后我需要添加一个或多个0以确保它的总数为4位

所以我需要它看起来像这样:

AFI12001,01,C-,201405,P,0430,2,0.02125000,US,60.0000
AFI12001,01,S-,201404,C,1550,2,0.03500000,US,30.0000

这段代码添加了0

import csv
new_rows = []
with open('csvpatpos.csv','r') as f:
csv_f = csv.reader(f)
for row in csv_f:
new_row = ""
col = 0
print row
for x in row:
col = col + 1
if col == 6:
if len(x) == 3:
x = "0" + x
new_row = new_row + x + ","
print new_row

然而，我在删除前面的零时遇到了麻烦。

Answer 1

将列转换为整型，然后再转换为任意格式的字符串

row[5] = "%04d" % int(row[5])

Answer 2

您可以在.lstrip()的几个步骤中完成此操作，然后找到结果字符串长度，然后将4-len(s) 0添加到前面。然而，我认为regex更容易。

with open('infilename', 'r') as infile:
    reader = csv.reader(infile)
    for row in reader:
        stripped_value = re.sub(r'^0{3}', '', row[5])

收益率

0430
1550

在正则表达式中，我们使用sub(pattern, substitute, original)格式。模式分解为:

'^' - match start of string
'0{3}' - match 3 zeros

您说第6列中的所有字符串都有7位数字，而您想要4位，因此将前3位替换为空字符串。

---

编辑:如果你想替换行，我就把它写到一个新文件:

with open('infilename', 'r') as infile, open('outfilename', 'w') as outfile:
    reader = csv.reader(infile)
    writer = csv.writer(outfile)
    for row in reader:
        row[5] = re.sub(r'^0{3}', '', row[5])
        writer.writerow(row)

---

Edit2:根据您最新的请求，我建议您这样做:

with open('infilename', 'r') as infile, open('outfilename', 'w') as outfile:
    reader = csv.reader(infile)
    writer = csv.writer(outfile)
    for row in reader:
        # strip all 0's from the front
        stripped_value = re.sub(r'^0+', '', row[5])
        # pad zeros on the left to smaller numbers to make them 4 digits
        row[5] = '%04d'%int(stripped_value)
        writer.writerow(row)

给定下列数字，

['0000430', '0001550', '0013300', '0012900', '0100000', '0001000']

这个收益率

['0430', '1550', '13300', '12900', '100000', '1000']

Answer 3

可以使用lstrip()和zfill()方法。这样的:

with open('input') as in_file:
    csv_reader = csv.reader(in_file)
    for row in csv_reader:
        stripped_data = row[5].lstrip('0')
        new_data = stripped_data.zfill(4)
        print new_data

这个打印

:

0430
1550

行:

stripped_data = row[5].lstrip('0')

去掉左边所有的0。这一行:

new_data = stripped_data.zfill(4) 

用0填充前面，使数字总数为4。

Answer 4

可以保留最后4个字符

columns[5] = columns[5][-4:]

例子

data = '''AFI12001,01,C-,201405,P,0000430,2,0.02125000,US,60.0000
AFI12001,01,S-,201404,C,0001550,2,0.03500000,US,30.0000'''
for row in data.splitlines():
    columns = row.split(',')
    columns[5] = columns[5][-4:]
    print ','.join(columns)

结果

AFI12001,01,C-,201405,P,0430,2,0.02125000,US,60.0000
AFI12001,01,S-,201404,C,1550,2,0.03500000,US,30.0000

---

编辑:

代码与csv模块-而不是data模拟文件。

import csv
with open('csvpatpos.csv','r') as f:
    csv_f = csv.reader(f)
    for row in csv_f:
        row[5] = row[5][-4:]
        print row[5] # print one element
        #print ','.join(row) # print full row
        print row # print full row