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    <pre><code>import pandas as pd def promotion(ls): return (ls.diff() &gt; 0).sum() def growth(ls): return ls.iloc[-1] - ls.iloc[0] jobData = pd.DataFrame( {'candidate_id': [1, 2, 2, 2], 'TitleLevel': [2, 1, 2, 1]}) grouped = jobData.groupby("candidate_id") titlePromotion = grouped["TitleLevel"].agg(promotion) print(titlePromotion) # candidate_id # 1 0 # 2 1 # dtype: int64 titleGrowth = grouped["TitleLevel"].agg(growth) print(titleGrowth) # candidate_id # 1 0 # 2 0 # dtype: int64 </code></pre> <hr> <p>Some tips:</p> <p>If you define the generic function</p> <pre><code>def foo(ls): print(type(ls)) </code></pre> <p>and call</p> <pre><code>jobData.groupby("candidate_id")["TitleLevel"].apply(foo) </code></pre> <p>Python will print</p> <pre><code>&lt;class 'pandas.core.series.Series'&gt; </code></pre> <p>This is a low-brow but effective way to discover that calling <code>jobData.groupby(...)[...].apply(foo)</code> passes a <code>Series</code> to <code>foo</code>.</p> <hr> <p>The <code>apply</code> method calls <code>foo</code> once for every group. It can return a Series or a DataFrame with the resulting chunks glued together. It is possible to use <code>apply</code> when <code>foo</code> returns an object such as a numerical value or string, but in such cases I think using <code>agg</code> is preferred. A typical use case for using <code>apply</code> is when you want to, say, square every value in a group and thus need to return a new group of the same shape. </p> <p>The <code>transform</code> method is also useful in this situation -- when you want to <em>transform</em> every value in the group and thus need to return something of the same shape -- but the result can be different than that with <code>apply</code> since a different object may be passed to <code>foo</code> (for example, each column of a grouped dataframe would be passed to <code>foo</code> when using <code>transform</code>, while the entire group would be passed to <code>foo</code> when using <code>apply</code>. The easiest way to understand this is to experiment with a simple dataframe and the generic <code>foo</code>.)</p> <p>The <code>agg</code> method calls <code>foo</code> once for every group, but unlike <code>apply</code> it should return a single number per group. The group is <em>aggregated</em> into a value. A typical use case for using <code>agg</code> is when you want to count the number of items in the group. </p> <hr> <p>You can debug and understand what went wrong with your original code by using the generic <code>foo</code> function:</p> <pre><code>In [30]: grouped['TitleLevel'].apply(foo) 0 2 Name: 1, dtype: int64 -------------------------------------------------------------------------------- 1 1 2 2 3 1 Name: 2, dtype: int64 -------------------------------------------------------------------------------- Out[30]: candidate_id 1 None 2 None dtype: object </code></pre> <p>This shows you the Series that are being passed to <code>foo</code>. Notice that in the second Series, then index values are 1 and 2. So <code>ls[0]</code> raises a <code>KeyError</code>, since there is no label with value <code>0</code> in the second Series. </p> <p>What you really want is the first item in the Series. That is what <code>iloc</code> is for. </p> <p>So to summarize, use <code>ls[label]</code> to select the row of a Series with index value of <code>label</code>. Use <code>ls.iloc[n]</code> to select the <code>n</code>th row of the Series.</p> <p>Thus, to fix your code with a the least amount of change, you could use</p> <pre><code>def promotion(ls): pro =0 if len(ls)&gt;1: for j in range(1,len(ls)): if ls.iloc[j]&gt;ls.iloc[j-1]: pro += 1 return pro def growth(ls): head= ls.iloc[0] tail= ls.iloc[len(ls)-1] gro= tail-head return gro </code></pre>
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    1. COThank you soooo much! You gave a very detailed explanation. But can you do me an another favor to explain why my code runs wrong? Thank you so much!
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