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<p>我有一个包含员工工资信息的数据框。大约有90多万行。在</p>
<p>样品:</p>
<pre><code>+----+-------------+---------------+----------+
| | table_num | name | salary |
|----+-------------+---------------+----------|
| 0 | 001234 | John Johnson | 1200 |
| 1 | 001234 | John Johnson | 1000 |
| 2 | 001235 | John Johnson | 1000 |
| 3 | 001235 | John Johnson | 1200 |
| 4 | 001235 | John Johnson | 1000 |
| 5 | 001235 | Steve Stevens | 1000 |
| 6 | 001236 | Steve Stevens | 1200 |
| 7 | 001236 | Steve Stevens | 1200 |
| 8 | 001236 | Steve Stevens | 1200 |
+----+-------------+---------------+----------+
</code></pre>
<p>数据类型:</p>
^{pr2}$
<p>我需要添加一列有关加薪/降薪的信息。
我使用<code>shift()</code>函数比较行中的值。在</p>
<p>主要问题在于对整个数据集中所有唯一的雇员进行过滤和迭代。在</p>
<p><strong>在我的脚本中大约需要3个半小时。在</p>
<p>怎样才能更快?</strong></p>
<p>我的剧本:</p>
<pre><code># giving us only unique combination of 'table_num' and 'name'
# since there can be same 'table_num' for different 'name'
# and same names with different 'table_num' appears sometimes
names_df = df[['table_num', 'name']].drop_duplicates()
# then extracting particular name and table_num from Series
for i in range(len(names_df)): ### Bottleneck of whole script ###
t = names_df.iloc[i,[0,1]][0]
n = names_df.iloc[i,[0,1]][1]
# using shift() and lambda to check if there difference between two rows
diff_sal = (df[(df['table_num']==t)
& ((df['name']==n))]['salary'] - df[(df['table_num']==t)
& ((df['name']==n))]['salary'].shift(1)).apply(lambda x: 1 if x>0 else (-1 if x<0 else 0))
df.loc[diff_sal.index, 'inc'] = diff_sal.values
</code></pre>
<p>输入数据示例:</p>
<pre><code>df = pd.DataFrame({'table_num': ['001234','001234','001235','001235','001235','001235','001236','001236','001236'],
'name': ['John Johnson','John Johnson','John Johnson','John Johnson','John Johnson', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens', 'Steve Stevens'],
'salary':[1200.,1000.,1000.,1200.,1000.,1000.,1200.,1200.,1200.]})
</code></pre>
<p>样本输出:</p>
<pre><code>+----+-------------+---------------+----------+-------+
| | table_num | name | salary | inc |
|----+-------------+---------------+----------+-------|
| 0 | 001234 | John Johnson | 1200 | 0 |
| 1 | 001234 | John Johnson | 1000 | -1 |
| 2 | 001235 | John Johnson | 1000 | 0 |
| 3 | 001235 | John Johnson | 1200 | 1 |
| 4 | 001235 | John Johnson | 1000 | -1 |
| 5 | 001235 | Steve Stevens | 1000 | 0 |
| 6 | 001236 | Steve Stevens | 1200 | 0 |
| 7 | 001236 | Steve Stevens | 1200 | 0 |
| 8 | 001236 | Steve Stevens | 1200 | 0 |
+----+-------------+---------------+----------+-------+
</code></pre>