[英]How to create a pandas dataframe from a file with no common separator?
我有路由表數據,看起來像這樣:
Valid Network Next-Hop Path Protocol
0 1.0.128.0/17 80.249.208.85 3257 38040 9737 i
0 80.249.209.150 6939 4766 38040 9737 i
1 80.249.209.37 3491 38040 9737 i
0 80.249.211.42 6762 38040 9737 i
0 80.249.208.85 3257 38040 9737 i
1 80.249.209.37 3491 38040 9737 i
0 80.249.211.42 6762 38040 9737 i
我想使用與列和網絡列中的前綴相同的列名稱創建DataFrame。 這里的問題是,並非所有行都有前綴,因此我需要添加最新的前綴(最近看到的)。 這是我所做的:
f = open('initial_data')
current_prefix = None
for index,line in enumerate(f):
if index != 0 and index != 1058274: #ignoring first and last line
if line.split(' ')[2].startswith(str(1)) or line.split(' ')[2].startswith(str(2)):
current_prefix = np.asarray(line.split(' ')[2]) #storing the most recent prefix
#print(current_prefix)#.shape,type(current_prefix))
df2 = pd.DataFrame([[current_prefix]], columns=list('Network'))
df.append(df2,ignore_index = True)#current_prefix)
else:
pass#df['Network'].append(current_prefix)
df2 = pd.DataFrame([[current_prefix]], columns=list('Network'))
df.append(df2,ignore_index = True)#current_prefix
但是前綴(例如1.0.128.0/17)被解釋為具有7列,並且出現此錯誤:
---------------------------------------------------------------------------
AssertionError Traceback (most recent call last)
<ipython-input-100-f2ee3d75b5c4> in <module>()
6 current_prefix = np.asarray(line.split(' ')[2])
7 #print(current_prefix)#.shape,type(current_prefix))
----> 8 df2 = pd.DataFrame([[current_prefix]], columns=list('Network'))
9 df.append(df2,ignore_index = True)#current_prefix)
AssertionError: 7 columns passed, passed data had 1 columns
那么,有沒有更好/更清潔的方法來解決這個問題? 更准確地說,我希望DataFrame看起來像這樣:
Valid | Network | Next-Hop | Path | Protocol
0 | 1.0.128.0/17 | 80.249.208.85 | 3257 38040 9737 | i
0 |NaN/aboveprefix | 80.249.209.150| 6939 4766 38040 9737 | i
等等。 有線索嗎?
我首先用\\ t分隔符編寫了一個新文件,然后使用pandas.read_csv('file_name',sep ='\\ t')解決了這個問題。
這里的主要技巧是計算每一行中的“ /”和“。”:
def classify(strx):
#call this on each line
next_hop = 0
prefix = ""
#path = ""
path = []
for i in strx.split(' '):
slash_counter,dot_counter = i.count('/'),i.count('.')
#print(i.count('.'),i.count('/'))
if dot_counter == 3 and slash_counter == 1:
prefix = i
elif dot_counter == 3 and slash_counter == 0:
next_hop = i
elif len(i) > 1 and dot_counter == 0 and slash_counter == 0:
#path = path.join(i)
path.append(i)
#print("Sanity check on paths",path,"\t",i)
#path = path.join(' ')
path_ = path
path = " ".join([str(i) for i in path_])
protocol = strx[-1]
#print(protocol)
#print("prefix = {0}, next hop = {1}, path = {2}".format(prefix,next_hop,path))
return(prefix,next_hop,path,protocol)
一個例子:原始行:
'0 1.0.128.0/17 80.249.208.85 3257 38040 9737 i'
用上面的函數轉換后:
1.0.128.0/17 80.249.208.85 3257 38040 9737我
接受的答案當然是實現此目的的一種方法,但是pandas
提供了一種提取字段並完全按照您所描述的方法通過將regex傳遞給Series
對象的str.extract()
來創建DataFrame
。然后使用method='ffill'
.fillna()
填寫“網絡”字段中的缺失值。 這種方法將通過以下類似方式完成。
import io
import pandas as pd
import re
f = io.StringIO('''\
Valid Network Next-Hop Path Protocol
0 1.0.128.0/17 80.249.208.85 3257 38040 9737 i
0 80.249.209.150 6939 4766 38040 9737 i
1 80.249.209.37 3491 38040 9737 i
0 80.249.211.42 6762 38040 9737 i
0 80.249.208.85 3257 38040 9737 i
1 80.249.209.37 3491 38040 9737 i
0 80.249.211.42 6762 38040 9737 i
''')
pattern = re.compile(r'^(?P<valid>[01])'
r'\s+(?P<network>\d+\.\d+\.\d+\.\d+/\d+)?'
r'\s+(?P<next_hop>\d+\.\d+\.\d+\.\d+)'
r'\s+(?P<path>(?:\d+ )+)'
r'(?P<protocol>[a-z])$')
next(f) #skip the first line
df = (pd.Series(f.readlines())
.str.extract(pattern, expand=False)
.fillna(method='ffill'))
這導致看起來像一個DataFrame
。
Out [26]:
valid network next_hop path protocol
0 0 1.0.128.0/17 80.249.208.85 3257 38040 9737 i
1 0 1.0.128.0/17 80.249.209.150 6939 4766 38040 9737 i
2 1 1.0.128.0/17 80.249.209.37 3491 38040 9737 i
3 0 1.0.128.0/17 80.249.211.42 6762 38040 9737 i
4 0 1.0.128.0/17 80.249.208.85 3257 38040 9737 i
5 1 1.0.128.0/17 80.249.209.37 3491 38040 9737 i
6 0 1.0.128.0/17 80.249.211.42 6762 38040 9737 i
如果您不希望填寫缺少的“網絡”值,則可以刪除對.fillna()
的調用。
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