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2019-02-22 04:16发布
我欲成王,谁敢阻挡
Screenshot of the query below:
Is there a way to easily drop the upper level column index and a have a single level with labels such as points_prev_amax, points_prev_amin, gf_prev_amax, gf_prev_amin and so on?
points_prev_amax
points_prev_amin
gf_prev_amax
gf_prev_amin
Using @jezrael's setup
df = pd.DataFrame({'A':[1,2,2,1], 'B':[4,5,6,4], 'C':[7,8,9,1], 'D':[1,3,5,9]}) df = df.groupby('A').agg([max, min])
Assign new columns with
from itertools import starmap def flat(midx, sep=''): fstr = sep.join(['{}'] * midx.nlevels) return pd.Index(starmap(fstr.format, midx)) df.columns = flat(df.columns, '_') df
Use list comprehension for set new column names:
list comprehension
df.columns = df.columns.map('_'.join) Or: df.columns = ['_'.join(col) for col in df.columns]
Sample:
df = pd.DataFrame({'A':[1,2,2,1], 'B':[4,5,6,4], 'C':[7,8,9,1], 'D':[1,3,5,9]}) print (df) A B C D 0 1 4 7 1 1 2 5 8 3 2 2 6 9 5 3 1 4 1 9 df = df.groupby('A').agg([max, min]) df.columns = df.columns.map('_'.join) print (df) B_max B_min C_max C_min D_max D_min A 1 4 4 7 1 9 1 2 6 5 9 8 5 3
print (['_'.join(col) for col in df.columns]) ['B_max', 'B_min', 'C_max', 'C_min', 'D_max', 'D_min'] df.columns = ['_'.join(col) for col in df.columns] print (df) B_max B_min C_max C_min D_max D_min A 1 4 4 7 1 9 1 2 6 5 9 8 5 3
If need prefix simple swap items of tuples:
prefix
df.columns = ['_'.join((col[1], col[0])) for col in df.columns] print (df) max_B min_B max_C min_C max_D min_D A 1 4 4 7 1 9 1 2 6 5 9 8 5 3
Another solution:
df.columns = ['{}_{}'.format(i[1], i[0]) for i in df.columns] print (df) max_B min_B max_C min_C max_D min_D A 1 4 4 7 1 9 1 2 6 5 9 8 5 3
If len of columns is big (10^6), then rather use to_series and str.join:
len
to_series
str.join
df.columns = df.columns.to_series().str.join('_')
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Using @jezrael's setup
Assign new columns with
Use
list comprehension
for set new column names:Sample:
If need
prefix
simple swap items of tuples:Another solution:
If
len
of columns is big (10^6), then rather useto_series
andstr.join
: