How do I robustly parse malformed CSV?

Question

I\'m processing data from government sources (FEC, state voter databases, etc). It\'s inconsistently malformed, which breaks my CSV parser in all sorts of delightful ways.

Answer 1

I made an app to reformat CSV files, doubling the single quotes inside fields and replacing the new lines inside them with a string like '\n'.

Once the data is inside the database we can replace back the '\n' to new lines.

I needed to do this because the apps I had to process CSV does not deal correctly with new lines.

Feel free to use and change.

In python:

import sys

def ProcessCSV(filename):
    file1 = open(filename, 'r')
    filename2 = filename + '.out'
    file2 = open(filename2, 'w')
    print 'Reformatting {0} to {1}...', filename, filename2
    line1 = file1.readline()
    while (len(line1) > 0):
        line1 = line1.rstrip('\r\n')
        line2 = ''
        count = 0
        lastField = ( len(line1) == 0 )
        while not lastField:
            lastField = (line1.find('","') == -1)
            res = line1.partition('","')
            field = res[0]
            line1 = res[2]
            count = count + 1
            hasStart = False
            hasEnd = False

            if  ( count == 1 )  and  ( field[:1] == '"' ) :
                field = field[1:]
                hasStart = True
            elif count > 1:
                hasStart = True

            while (True):
                if  ( lastField == True )  and  ( field[-1:] == '"' ) :
                    field = field[:-1]
                    hasEnd = True
                elif not lastField:
                    hasEnd = True

                if lastField and not hasEnd:
                    line1 = file1.readline()
                    if (len(line1) == 0): break
                    line1 = line1.rstrip('\r\n')
                    lastField = (line1.find('","') == -1)
                    res = line1.partition('","')
                    field = field + '\\n' + res[0]
                    line1 = res[2]
                else:
                    break

            field = field.replace('"', '""')

            line2 = line2 + iif(count > 1, ',', '') + iif(hasStart, '"', '') + field + iif(hasEnd, '"', '')

        if len(line2) > 0:
            file2.write(line2)
            file2.write('\n')

        line1 = file1.readline()

    file1.close()
    file2.close()
    print 'Done'

def iif(st, v1, v2):
    if st:
        return v1
    else:
        return v2

filename = sys.argv[1]
if len(filename) == 0:
    print 'You must specify the input file'
else:
    ProcessCSV(filename)

In VB.net:

Module Module1

Sub Main()
    Dim FileName As String
    FileName = Command()
    If FileName.Length = 0 Then
        Console.WriteLine("You must specify the input file")
    Else
        ProcessCSV(FileName)
    End If
End Sub

Sub ProcessCSV(ByVal FileName As String)
    Dim File1 As Integer, File2 As Integer
    Dim Line1 As String, Line2 As String
    Dim Field As String, Count As Long
    Dim HasStart As Boolean, HasEnd As Boolean
    Dim FileName2 As String, LastField As Boolean
    On Error GoTo locError

    File1 = FreeFile()
    FileOpen(File1, FileName, OpenMode.Input, OpenAccess.Read)

    FileName2 = FileName & ".out"
    File2 = FreeFile()
    FileOpen(File2, FileName2, OpenMode.Output)

    Console.WriteLine("Reformatting {0} to {1}...", FileName, FileName2)

    Do Until EOF(File1)
        Line1 = LineInput(File1)
        '
        Line2 = ""
        Count = 0
        LastField = (Len(Line1) = 0)
        Do Until LastField
            LastField = (InStr(Line1, """,""") = 0)
            Field = Strip(Line1, """,""")
            Count = Count + 1
            HasStart = False
            HasEnd = False
            '
            If (Count = 1) And (Left$(Field, 1) = """") Then
                Field = Mid$(Field, 2)
                HasStart = True
            ElseIf Count > 1 Then
                HasStart = True
            End If
            '
locFinal:
            If (LastField) And (Right$(Field, 1) = """") Then
                Field = Left$(Field, Len(Field) - 1)
                HasEnd = True
            ElseIf Not LastField Then
                HasEnd = True
            End If
            '
            If LastField And Not HasEnd And Not EOF(File1) Then
                Line1 = LineInput(File1)
                LastField = (InStr(Line1, """,""") = 0)
                Field = Field & "\n" & Strip(Line1, """,""")
                GoTo locFinal
            End If
            '
            Field = Replace(Field, """", """""")
            '
            Line2 = Line2 & IIf(Count > 1, ",", "") & IIf(HasStart, """", "") & Field & IIf(HasEnd, """", "")
        Loop
        '
        If Len(Line2) > 0 Then
            PrintLine(File2, Line2)
        End If
    Loop

    FileClose(File1, File2)
    Console.WriteLine("Done")

    Exit Sub
locError:
    Console.WriteLine("Error: " & Err.Description)
End Sub

Function Strip(ByRef Text As String, ByRef Separator As String) As String
    Dim nPos As Long
    nPos = InStr(Text, Separator)
    If nPos > 0 Then
        Strip = Left$(Text, nPos - 1)
        Text = Mid$(Text, nPos + Len(Separator))
    Else
        Strip = Text
        Text = ""
    End If
End Function

End Module

Answer 2

First, here is a rather naive attempt: http://rubular.com/r/gvh3BJaNTc

/"(.*?)"(?=[\r\n,]|$)|([^,"\s].*?)(?=[\r\n,]|$)/m

The assumptions here are:

• A field may start with quotes. In which case, it should end with a quote that is either:
  • before a comma
  • before a new line (if it is last field on its line)
  • before the end of the file (if it is last field on the last line)
• Or, its first character is not a quote, so it contains characters until the same condition as before is met.

This almost does what you want, but fails on these fields:

1 comma and
linebreaks"

As TC had pointed out in the comments, your text is ambiguous. I'm sure you already know it, but for completeness:

• "a" - is that a or "a"? How do you represent a value that you want to be wrapped in quotes?
• "1","2" - might be parsed as 1,2, or as 1","2 - both are legal.
• ,1 \n 2, - End of line, or newline in the value? You cannot tell, specially if this is supposed to be the last value of its line.
• 1 \n 2 \n 3 - One value with newlines? Two values (1\n2,3 or 1,2\n3)? Three values?

You may be able to get some clues if you examine the first value on each row, which as you have said, should tell you the number of columns and their types - this can give you the additional information you are missing to parse the file (for example, if you know there should another field in this line, then all newlines belong in the current value). Even then though, it looks like there are serious problems here...

Answer 3

It is possible to subclass Ruby's File to process each line of the the CSV file before it is passed to the Ruby's CSV parser. For example, here's how I used this trick to replace non-standard backslash-escaped quotes \" with standard double-quotes ""

class MyFile < File
  def gets(*args)
    line = super
    if line != nil
      line.gsub!('\\"','""')  # fix the \" that would otherwise cause a parse error
    end
    line
  end
end

infile = MyFile.open(filename)
incsv = CSV.new(infile)

while row = incsv.shift
  # process each row here
end

You could in principle do all sorts of additional processing, e.g. UTF-8 cleanups. The nice thing about this approach is you handle the file on a line by line basis, so you don't need to load it all into memory or create an intermediate file.