使用python和pandas将CSV文件中的三列合并为一列
嗨,我正在尝试将几个现有列合并成一个新列,然后删除CSV文件中的三个原始列。我一直试图用熊猫来做这件事,但是运气不太好。我对python相当陌生 我的代码首先将几个CSV文件合并到同一个目录中,然后尝试操作列。第一次合并有效,我得到一个带有合并数据的output.csv,但是列的合并无效使用python和pandas将CSV文件中的三列合并为一列,python,pandas,csv,dataframe,Python,Pandas,Csv,Dataframe,嗨,我正在尝试将几个现有列合并成一个新列,然后删除CSV文件中的三个原始列。我一直试图用熊猫来做这件事,但是运气不太好。我对python相当陌生 我的代码首先将几个CSV文件合并到同一个目录中,然后尝试操作列。第一次合并有效,我得到一个带有合并数据的output.csv,但是列的合并无效 import glob import pandas as pd interesting_files = glob.glob("*.csv") header_saved = False with open('
import glob
import pandas as pd
interesting_files = glob.glob("*.csv")
header_saved = False
with open('output.csv','wb') as fout:
for filename in interesting_files:
with open(filename) as fin:
header = next(fin)
if not header_saved:
fout.write(header)
header_saved = True
for line in fin:
fout.write(line)
df = pd.read_csv("output.csv")
df['HostAffected']=df['Host'] + "/" + df['Protocol'] + "/" + df['Port']
df.to_csv("newoutput.csv")
有效地扭转这一局面:
Host,Protocol,Port
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,49707
10.0.0.10,tcp,49672
10.0.0.10,tcp,49670
变成这样:
HostsAffected
10.0.0.10/tcp/445
10.0.0.10/tcp/445
10.0.0.10/tcp/445
10.0.0.10/tcp/445
10.0.0.10/tcp/445
10.0.0.10/tcp/445
10.0.0.11/tcp/445
10.0.0.11/tcp/49707
10.0.0.11/tcp/49672
10.0.0.11/tcp/49670
10.0.0.11/tcp/49668
10.0.0.11/tcp/49667
但是,csv中还有其他列
我不是编码员,我只是想解决一个问题,非常感谢您的帮助。这就是您可以做到的:
dt = """Host,Protocol,Port
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,49707
10.0.0.10,tcp,49672
10.0.0.10,tcp,49670"""
tdf = pd.read_csv(pd.compat.StringIO(dt))
tdf['HostsAffected'] = tdf.apply(lambda x: '{}/{}/{}'.format(x['Host'] , x['Protocol'] , x['Port']), axis=1)
tdf = tdf[['HostsAffected']]
tdf.to_csv(<path-to-save-csv-file>)
如果要从文件中读取CSV,请按如下方式编辑读取CSV行:
tdf = pd.read_csv(<path-to-the-file>)
tdf=pd.read\u csv()
有两种方法可以做到这一点:要么使用矢量化函数组合系列,要么使用lambda
函数与pd.series.apply
矢量化解决方案
df['HostsAffected'] = df.apply(lambda x: '/'.join(list(map(str, x))), axis=1)
不要忘记将非数字类型转换为str
df['HostAffected'] = df['Host'] + '/' + df['Protocol'] + '/' + df['Port'].map(str)
业绩说明:
应用lambda
功能
df['HostsAffected'] = df.apply(lambda x: '/'.join(list(map(str, x))), axis=1)
对于这两种解决方案,您只需按此列筛选即可删除所有其他解决方案:
df = df[['HostsAffected']]
完整示例
from io import StringIO
import pandas as pd
mystr = StringIO("""Host,Protocol,Port
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,445
10.0.0.10,tcp,49707
10.0.0.10,tcp,49672
10.0.0.10,tcp,49670""")
# replace mystr with 'file.csv'
df = pd.read_csv(mystr)
# combine columns
df['HostsAffected'] = df['Host'] + '/' + df['Protocol'] + '/' + df['Port'].map(str)
# include only new columns
df = df[['HostsAffected']]
结果:
print(df)
HostsAffected
0 10.0.0.10/tcp/445
1 10.0.0.10/tcp/445
2 10.0.0.10/tcp/445
3 10.0.0.10/tcp/445
4 10.0.0.10/tcp/445
5 10.0.0.10/tcp/445
6 10.0.0.10/tcp/445
7 10.0.0.10/tcp/49707
8 10.0.0.10/tcp/49672
9 10.0.0.10/tcp/49670
在我看来,我们有三种选择:
%timeit df['Host'] + "/" + df['Protocol'] + "/" + df['Port'].map(str)
%timeit ['/'.join(i) for i in zip(df['Host'],df['Protocol'],df['Port'].map(str))]
%timeit ['/'.join(i) for i in df[['Host','Protocol','Port']].astype(str).values]
计时:
10 loops, best of 3: 39.7 ms per loop
10 loops, best of 3: 35.9 ms per loop
10 loops, best of 3: 162 ms per loop
然而,我认为这是你最容易理解的方法:
import pandas as pd
data = '''\
ID,Host,Protocol,Port
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,445
1,10.0.0.10,tcp,49707
1,10.0.0.10,tcp,49672
1,10.0.0.10,tcp,49670'''
df = pd.read_csv(pd.compat.StringIO(data)) # Recreates a sample dataframe
cols = ['Host','Protocol','Port']
newcol = ['/'.join(i) for i in df[cols].astype(str).values]
df = df.assign(HostAffected=newcol).drop(cols, 1)
print(df)
返回:
ID HostAffected
0 1 10.0.0.10/tcp/445
1 1 10.0.0.10/tcp/445
2 1 10.0.0.10/tcp/445
3 1 10.0.0.10/tcp/445
4 1 10.0.0.10/tcp/445
5 1 10.0.0.10/tcp/445
6 1 10.0.0.10/tcp/445
7 1 10.0.0.10/tcp/49707
8 1 10.0.0.10/tcp/49672
9 1 10.0.0.10/tcp/49670
很抱歉,问题出在哪里?可能是复制品,但解决速度很慢,最好是原创的-看你是对的。我听到很多人抱怨
apply
@jezrael不应该这样:df['hostimpacted']=['/')。在zip中加入(I)for I(df['Host']、df['Protocol']、df['Port'].astype(str))
是最快的吗?@AntonvBR-最好的测试,不确定。@AntonvBR nice one antov.谢谢你在这方面的帮助。将.map(str)添加到合并列的末尾修复了该问题@jpp好的。现在已经证实了zip()
应该是最快的解决方案。是的,我同意。谢谢你的更新。更接近,但现在我们有了很好的基准。我也尝试过这种方法,效果很好。谢谢。@Anton vBR如果这些字段中有一个为空,脚本就会停止-有办法解决吗?例如,如果tcp在任何给定的行中丢失,它将在哪里停止,因为它应该跳过并合并ip和端口。@Abob在'
中为空?