假设我有
edu_data = [['school', 1, 2], ['college', 3, 4], ['grad-school', 5, 6]]
edu = pd.DataFrame(edu_data, columns = ['Education', 'StudentID1', 'StudentID2'])
print(edu)
Education StudentID1 StudentID2
0 school 1 2
1 college 3 4
2 grad-school 5 6
然后我还有另一个带有学生证的表：
data = [['tom', 3], ['nick', 5], ['juli', 6], ['jack', 10]]
df = pd.DataFrame(data, columns = ['Name', 'StudentID'])
print(df)
Name StudentID
0 tom 3
1 nick 5
2 juli 6
3 jack 10
如何获得一个表，将df ['StudentID']与edu [“ StudentID1”]或edu [“ StudentID2”]进行匹配。如果df ['StudentID']等于任一，那么我想将edu [“ Education”]附加到df。
所以我希望我的输出是：
Name StudentID Education
0 tom 3 college
1 nick 5 grad-school
2 juli 6 grad-school
3 jack 10 NaN
答案 0 :(得分：2)
您可以使用DataFrame.melt：
mapper=edu.melt(id_vars='Education',var_name = 'StudentID',value_name='ID').set_index('ID')
df['Education']=df['StudentID'].map(mapper['Education'])
print(df)
Name StudentID Education
0 tom 3 college
1 nick 5 grad-school
2 juli 6 grad-school
3 jack 10 NaN
详细信息：
print(mapper)
Education StudentID
ID
1 school StudentID1
3 college StudentID1
5 grad-school StudentID1
2 school StudentID2
4 college StudentID2
6 grad-school StudentID2
您还可以使用Series.map + Series.combine_first：
eduID1=edu.set_index('StudentID1')
eduID2=edu.set_index('StudentID2')
df['Education']=df['StudentID'].map(eduID1['Education']).combine_first(df['StudentID'].map(eduID2['Education']))
print(df)
Name StudentID Education
0 tom 3 college
1 nick 5 grad-school
2 juli 6 grad-school
3 jack 10 NaN
答案 1 :(得分：2)
使用@extends('layouts.app')
@section('content')
<div class="row">
<div class="col-sm-8 offset-sm-2">
<h1 class="display-3">Update a Stock</h1>
@if ($errors->any())
<div class="alert alert-danger">
<ul>
@foreach ($errors->all() as $error)
<li>{{ $error }}</li>
@endforeach
</ul>
</div>
<br />
@endif
<form method="post" action="{{ route('updateStock', $Stock->id) }}">
{{ csrf_field() }}
<div class="form-group">
<label for="stock_name">Stock Name:</label>
<input type="text" class="form-control" name="stock_name" value={{$Stock->stock_name }} />
</div>
<div class="form-group">
<label for="stock_qty">Stock Amount:</label>
<input type="number" class="form-control" name="stock_qty" value={{$Stock->stock_qty }} />
</div>
<div class="form-group">
<label for="stock_unit">Stock Unit:</label>
<select id="stock_unit" name="stock_unit" value={{$Stock->stock_unit}}>
<option value="Kg">Kg</option>
<option value="Qty">Qty</option>
</select>
</div>
<div class="form-group">
<label for="stock_price_per_kg">Price Per Kg:</label>
<input type="number" class="form-control" name="stock_price_per_kg" value={{$Stock->stock_price_per_kg }} />
</div>
<div class="form-group">
<label for="stock_weight_per_qty">Weight Per Qty:</label>
<input type="number" class="form-control" name="stock_weight_per_qty" value={{$Stock->stock_weight_per_qty }} />
</div>
<button type="submit" class="btn btn-primary">Update</button>
</form>
</div>
</div>
@endsection
maps = edu.set_index('Education').stack().reset_index(level=1, drop=True)
df['Education'] = df.StudentID.map(pd.Series(s.index, s.values))
答案 2 :(得分：1)
这是melt和merge的解决方案：
(df.merge(edu.melt(id_vars='Education',
value_name='StudentID'),
on='StudentID',
how='left')
.drop_duplicates(['Name','StudentID']) # this is for when both StudentID match, we choose the first
.drop('variable', axis=1)
)
输出：
Name StudentID Education
0 tom 3 college
1 nick 5 grad-school
2 juli 6 grad-school
3 jack 10 NaN
答案 3 :(得分：1)
使用地图，类似于今天早些时候我的回答
mapper = edu.set_index('StudentID1')['Education'].to_dict()
mapper.update(edu.set_index('StudentID2')['Education'].to_dict())
df['Education'] = df['StudentID'].map(mapper)
Name StudentID Education
0 tom 3 college
1 nick 5 grad-school
2 juli 6 grad-school
3 jack 10 NaN