Python duplicatesの例

コード例 #1

ファイル: examples.py プロジェクト: datamade/petl

look(table10)

# duplicates

table1 = [['foo', 'bar', 'baz'],
          ['A', 1, 2.0],
          ['B', 2, 3.4],
          ['D', 6, 9.3],
          ['B', 3, 7.8],
          ['B', 2, 12.3],
          ['E', None, 1.3],
          ['D', 4, 14.5]]

from petl import duplicates, look    
look(table1)
table2 = duplicates(table1, 'foo')
look(table2)
# compound keys are supported
table3 = duplicates(table1, key=['foo', 'bar'])
look(table3)
    

# conflicts

table1 = [['foo', 'bar', 'baz'],
          ['A', 1, 2.7],
          ['B', 2, None],
          ['D', 3, 9.4],
          ['B', None, 7.8],
          ['E', None],
          ['D', 3, 12.3],

コード例 #2

ファイル: examples.py プロジェクト: aklimchak/petl

look(table10)

# duplicates

table1 = [['foo', 'bar', 'baz'],
          ['A', 1, 2.0],
          ['B', 2, 3.4],
          ['D', 6, 9.3],
          ['B', 3, 7.8],
          ['B', 2, 12.3],
          ['E', None, 1.3],
          ['D', 4, 14.5]]

from petl import duplicates, look    
look(table1)
table2 = duplicates(table1, 'foo')
look(table2)
# compound keys are supported
table3 = duplicates(table1, key=['foo', 'bar'])
look(table3)
    

# conflicts

table1 = [['foo', 'bar', 'baz'],
          ['A', 1, 2.7],
          ['B', 2, None],
          ['D', 3, 9.4],
          ['B', None, 7.8],
          ['E', None],
          ['D', 3, 12.3],