python - 根据索引条件创建新列

Question

我有 df:

arrays = [np.array(['bar', 'bar', 'bar', 'bar', 'bar', 'bar', 'baz', 'baz', 'baz', 'baz', 'baz', 'baz', 'foo', 'foo', 'foo', 'foo', 'foo', 'foo']),
       np.array(['one','two'] * 9),
        np.array([1,2,3] * 6)]
df = pd.DataFrame(np.random.randn(18,2), index=arrays)


                      col0      col1
bar one     1   0.872359    -1.115871
    two     2   -0.937908   -0.528563
    one     3   -0.118874   0.286595
    two     1   -0.507698   1.364643
    one     2   1.507611    1.379498
    two     3   -1.398019   -1.603056
baz one     1   1.498263    0.412380
    two     2   -0.930022   -1.483657
    one     3   -0.438157   1.465089
    two     1   0.161887    1.346587
    one     2   0.167086    1.246322
    two     3   0.276344    -1.206415
foo one     1   -0.045389   -0.759927
    two     2   0.087999    -0.435753
    one     3   -0.232054   -2.221466
    two     1   -1.299483   1.697065
    one     2   0.612211    -1.076738
    two     3   -1.482573   0.907826

现在我想创建"new"列:

for 'bar'
  if index.level(2) > 1
     "NEW" = col1
  else
     "NEW" = col2

'baz' 与 >2 相同

'foo' 与 >3 相同

没有 Py 循环怎么办？

Answer 1

您可以使用 get_level_values用于按级别选择索引值，然后用于新列 numpy.where :

#if possible use dictionary
d = {'bar':1, 'baz':2, 'foo':3}
m = df.index.get_level_values(2)  > df.rename(d).index.get_level_values(0)

df['NEW'] = np.where(m, df.col1, df.col2)

对于更通用的解决方案，请使用 Series.rank :

a = df.index.get_level_values(2) 
b = df.index.get_level_values(0).to_series().rank(method='dense')

df['NEW'] = np.where(a > b, df.col1, df.col2)

详细信息:

print (b)
bar    1.0
bar    1.0
bar    1.0
bar    1.0
bar    1.0
bar    1.0
baz    2.0
baz    2.0
baz    2.0
baz    2.0
baz    2.0
baz    2.0
foo    3.0
foo    3.0
foo    3.0
foo    3.0
foo    3.0
foo    3.0
dtype: float64