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pythondelete-rowdbf

How to delete dbf rows containing an alpha value or an entire dbf column

I would like to delete some rows & columns on my dfb that contain specific data.

Let's assume that the structure of the dfb looks like the image bellow and that I want to remove the rows with 'knapford' as 'Prov' and the 'Reg' column.

enter image description here

When I try following this script:

table = dbf.Table(wonderland.dbf).open(mode=dbf.READ_WRITE)
with table as tab:
    print (table)
    for record in dbf.Process(tab):
        dbf.delete(record[0])

I get this error:

  File "<ipython-input-4-d8feabfc9819>", line 4, in <module>
    dbf.delete(record[0])

  File "C:\Users\fassi\AppData\Local\Continuum\anaconda3\lib\site-packages\dbf\__init__.py", line 8349, in delete
    if not template and record._meta.status == CLOSED:

AttributeError: 'int' object has no attribute '_meta'

I had this other suggestion in mind, but I wonder if it will really delete all the fields of the row, just some specific ones or all the records!

table = dbf.Table(wonderland.dbf).open(mode=dbf.READ_WRITE)
with table as tab:
    dbf.delete_fields(table, Reg)
    for record in dbf.Process(tab):
        if record[0].strip()=='Knapford': #since we've delete the Reg, Prov is becoming record[0]?
            dbf.delete(record)

Thanks in advance

Solution

  • I think that this should work:

    from simpledbf import Dbf5

dbf = Dbf5('file.dbf')
data = dbf.to_dataframe()

for row in range(len(data)):
    Reg= data.loc[row,'Reg']
    Prov = data.loc[row,'Prov']
    if Reg =='knapford' or Prov =='knapford':
        data.drop(index=row, axis=0, inplace=True)

    This other script work very well:

    import dbf
with table as tab:
    for record in dbf.Process(tab):
        if record[1].strip()=='Knapford':
            #print (record)
            dbf.delete(record)
table.pack()