<p>你可以像这样列举拉链</p>
<pre><code>dd = {'name': ["HARDIE'S MOBILE HOME PARK", 'CRESTVIEW RV PARK',
'HOMESTEAD TRAILER PARK', 'HOUSTON PARK MOBILE HOME PARK',
'HUDSON MOBILE HOME PARK', 'BEACH DRIVE MOBILE HOME PARK',
'EVANS TRAILER PARK'],
'country': ['USA', 'USA', 'USA', 'USA', 'USA', 'USA', 'USA'],
'coordinates': ['30.44126118, -86.6240656099999',
'30.7190163500001, -86.5716222299999',
'30.5115772500001, -86.4628417499999',
'30.4424195300001, -86.64733076',
'30.7629176200001, -86.5928893399999', '30.44417349, -86.59951996',
'30.4427800300001, -86.62941091'],
'status':['OPEN', 'CLOSED', 'OPEN', 'OPEN', 'OPEN', 'OPEN', 'OPEN']}
df1 = pd.DataFrame(data=dd)
d_out = {
f"destination{idx+1}":'; '.join(v) for idx, v in enumerate(zip(df1.name[1:], df1.coordinates[1:]))
}
d_out
{'destination1': 'CRESTVIEW RV PARK; 30.7190163500001, -86.5716222299999',
'destination2': 'HOMESTEAD TRAILER PARK; 30.5115772500001, -86.4628417499999',
'destination3': 'HOUSTON PARK MOBILE HOME PARK; 30.4424195300001, -86.64733076',
'destination4': 'HUDSON MOBILE HOME PARK; 30.7629176200001, -86.5928893399999',
'destination5': 'BEACH DRIVE MOBILE HOME PARK; 30.44417349, -86.59951996',
'destination6': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091'}
</code></pre>
<p>你不必做dict理解就能得到这个结果,如果你能像这样在pandas数据框中做几列,你就能得到这个结果</p>
<pre><code>df1['destination'] = [f"destination{k}" for k in range(len(df1))]
df1['value'] = df1['name'] + "; " + df1['coordinates']
df1[['destination', 'value']][1:].set_index("destination").to_dict()['value']
{'destination1': 'CRESTVIEW RV PARK; 30.7190163500001, -86.5716222299999',
'destination2': 'HOMESTEAD TRAILER PARK; 30.5115772500001, -86.4628417499999',
'destination3': 'HOUSTON PARK MOBILE HOME PARK; 30.4424195300001, -86.64733076',
'destination4': 'HUDSON MOBILE HOME PARK; 30.7629176200001, -86.5928893399999',
'destination5': 'BEACH DRIVE MOBILE HOME PARK; 30.44417349, -86.59951996',
'destination6': 'EVANS TRAILER PARK; 30.4427800300001, -86.62941091'}
</code></pre>