Pandas的apply函数返回多列结果

方法1

series是一个字典

用字典的方式,新增字段值;

这个字典中,将包含旧的值,以及新的值;

def sizes(s):
    s['size_kb'] = locale.format("%.1f", s['size'] / 1024.0, grouping=True) + ' KB'
    s['size_mb'] = locale.format("%.1f", s['size'] / 1024.0 ** 2, grouping=True) + ' MB'
    s['size_gb'] = locale.format("%.1f", s['size'] / 1024.0 ** 3, grouping=True) + ' GB'
    return s
df_test = df_test.append(rows_list)
df_test = df_test.apply(sizes, axis=1)

方法2:

拆解成三个字段列

def sizes(s):    
    return locale.format("%.1f", s / 1024.0, grouping=True) + ' KB', \
        locale.format("%.1f", s / 1024.0 ** 2, grouping=True) + ' MB', \
        locale.format("%.1f", s / 1024.0 ** 3, grouping=True) + ' GB'
df_test['size_kb'],  df_test['size_mb'], df_test['size_gb'] = zip(*df_test['size'].apply(sizes))

方法3:

import pandas as pd
df_test = pd.DataFrame([
  {'dir': '/Users/uname1', 'size': 994933},
  {'dir': '/Users/uname2', 'size': 109338711},
])
def sizes(s):
  a = locale.format_string("%.1f", s['size'] / 1024.0, grouping=True) + ' KB'
  b = locale.format_string("%.1f", s['size'] / 1024.0 ** 2, grouping=True) + ' MB'
  c = locale.format_string("%.1f", s['size'] / 1024.0 ** 3, grouping=True) + ' GB'
  return a, b, c
df_test[['size_kb', 'size_mb', 'size_gb']] = df_test.apply(sizes, axis=1, result_type="expand")

expand关键字的作用:

方法4:

def sizes(s):
    val_kb = locale.format("%.1f", s['size'] / 1024.0, grouping=True) + ' KB'
    val_mb = locale.format("%.1f", s['size'] / 1024.0 ** 2, grouping=True) + ' MB'
    val_gb = locale.format("%.1f", s['size'] / 1024.0 ** 3, grouping=True) + ' GB'
    return pd.Series([val_kb,val_mb,val_gb],index=['size_kb','size_mb','size_gb'])
df[['size_kb','size_mb','size_gb']] = df.apply(lambda x: sizes(x) , axis=1)

综合

import pandas as pd
dat = [ [i, 10*i] for i in range(1000)]
df = pd.DataFrame(dat, columns = ["a","b"])
def add_and_sub(row):
    add = row["a"] + row["b"]
    sub = row["a"] - row["b"]
    return add, sub
df[["add", "sub"]] = df.apply(add_and_sub, axis=1, result_type="expand")
# versus
df["add"], df["sub"] = zip(*df.apply(add_and_sub, axis=1))

来源:

https://stackoverflow.com/questions/23586510/return-multiple-columns-from-pandas-apply