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位数。用这个

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

我需要删除前面的零，然后我需要添加零或零以确保它总共有4位数

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

此代码添加了零

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

但是，我在删除前面的零时遇到问题。

Answer 1

将列转换为int，然后将其转换为您想要的任何格式的字符串。

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

Answer 2

您可以使用lstrip()和zfill()方法。像这样：

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

打印：

0430
1550

该行：

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

摆脱了左边的所有零。这一行：

new_data = stripped_data.zfill(4)

用零填充前面，使总位数为4.

希望这有帮助。

Answer 3

您可以在.lstrip()的几个步骤中执行此操作，然后找到生成的字符串长度，然后将4-len(s) 0添加到前面。但是，我认为使用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])

产量

0430
1550

在正则表达式中，我们使用格式sub(pattern, substitute, original)。模式细分是：

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

你说第6列中的所有字符串都有7个数字，你想要4，所以用空字符串替换前3个字符串。

编辑：如果您想要替换行，我只会将其写入新文件：

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 ：根据您的最新要求，我建议您执行以下操作：

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)

鉴于以下数字，

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

这会产生

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

Answer 4

你可以保留最后4个字符

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

例如

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：

使用csv模块的代码 - 而不是data来模拟文件。

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

使用python在csv中删除零

4 个答案: