python - Python Pandas：使用 apply 方法从另一个数据框中更新多个列值

Question

我想从另一个数据框中更新多列的值我使用 combine_first 找到了解决方案

'''

df1 = pd.DataFrame({'id' : ['1470', '1550', '1562', '1578', '1645', '1867', '1888', '2205', '2283', '2306'],
'gp' : ['nl_i', 'adv_i', 'adv_i', 'nl_i', 'adv_i', 'early_i', 'nl_i', 'nl_i', 'nl_i', 'nl_i'],
'cho' : [69626.0, 183425.0, 75418.0, 84239.0, 158721.0, 122857.0, 166052.0, 86686.0, 140407.0, 122792.0],
'date_i' : ['2000-11-29', '2000-11-28', '2000-11-27', '2000-11-26', '2000-11-25', '2000-11-24', '2000-11-23', '2000-11-22', '2000-11-21', '2000-11-20'],
})
df1['rl'] = np.NAN
df1['date_f'] = np.NAN

''' '''

df2 = pd.DataFrame({'no' : ['2939', '2283', '1578', '2781', '2319', '2306', '1888', '1470', '2869', '2205'],
'date_i' : ['2010-09-18', '2012-02-08', '2012-04-09', '2012-04-23', '2012-05-08', '2012-09-04', '2013-08-29', '2013-09-09', '2014-02-24', '2015-11-19'],
'rl'  : ['r', 'l', 'r', 'r', 'l', 'r', 'r', 'r', 'r', 'r']})

''' '''

df3 = pd.DataFrame ({'no' : ['2319', '2306', '1888', '1470', '2869', '2205'],
'date_f' : ['2019-05-10', '2013-09-24', '2019-06-12', '2016-08-29', '2016-10-10', '2017-11-30']})

'''

在这里，df1 的 'id' 与 df2 和 df3 的 'no' 中的值相同 我想从 df2 和 df3 的值更新 ['date_i', 'rl', 'date_f'] 列 下一个数据帧是我想要的

我使用 combine_first 在堆栈溢出中找到了这个解决方案

'''

df1_new = df2.set_index('no').combine_first(df1.set_index('id')).reset_index()
df1_new.rename(columns = {'index':'id'}, inplace = True)
df1_new = df1_new[df1.columns]
df1_new.dropna(subset=['cho'], inplace = True)

df1_new = df3.set_index('no').combine_first(df1_new.set_index('id')).reset_index()
df1_new.rename(columns = {'index':'id'}, inplace = True)
df1_new = df1_new[df1.columns]
df1_new.dropna(subset=['cho'], inplace = True)

'''

但是，我尝试使用 apply 方法解决这个问题

我可以使用 apply 方法而不是 combine_first 来更新我的数据框吗？

标签： pythonpandasapply

Answer 1

我们在不使用 apply 的情况下合并了每个数据框，替换了日期的 NaN 值并删除了不必要的列。如果你真的要使用apply函数，这个答案不会令人满意。

df1_new = df1.merge(df2, left_on='id', right_on='no', how='left')
df1_new = df1_new.merge(df3,left_on='id', right_on='no', how='left')
df1_new['date_i_y'].fillna('', inplace=True)

for idx, row in df1_new.iterrows():
    if len(row['date_i_y']) == 0:
        df1_new.loc[idx,'date_i_y'] = row['date_i_x']

df1_new.drop(['date_i_x','no_y'], axis=1, inplace=True)
df1_new.columns = ['id', 'gp', 'cho', 'no', 'date_i', 'rl', 'date_f']

df1_new
id  gp  cho     no  date_i  rl  date_f
0   1470    nl_i    69626.0     1470    2013-09-09  r   2016-08-29
1   1550    adv_i   183425.0    NaN     2000-11-28  NaN     NaN
2   1562    adv_i   75418.0     NaN     2000-11-27  NaN     NaN
3   1578    nl_i    84239.0     1578    2012-04-09  r   NaN
4   1645    adv_i   158721.0    NaN     2000-11-25  NaN     NaN
5   1867    early_i     122857.0    NaN     2000-11-24  NaN     NaN
6   1888    nl_i    166052.0    1888    2013-08-29  r   2019-06-12
7   2205    nl_i    86686.0     2205    2015-11-19  r   2017-11-30
8   2283    nl_i    140407.0    2283    2012-02-08  l   NaN
9   2306    nl_i    122792.0    2306    2012-09-04  r   2013-09-24