从元组到 Pandas 中的多列
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From tuples to multiple columns in pandas
提问by ba_ul
How do I convert this dataframe
我如何转换这个数据框
location value 0 (Richmond, Virginia, nan, USA) 100 1 (New York City, New York, nan, USA) 200
to this:
对此：
city state region country value 0 Richmond Virginia nan USA 100 1 New York City New York nan USA 200
Note that the locationcolumn in the first dataframe contains tuples. I want to create four columns out of the locationcolumn.
请注意，location第一个数据框中的列包含元组。我想从列中创建四列location。
采纳答案by exp1orer
new_col_list = ['city','state','regions','country']
for n,col in enumerate(new_col_list):
df[col] = df['location'].apply(lambda location: location[n])
df = df.drop('location',axis=1)
回答by meloncholy
If you return a Series of the (split) location, you can merge (jointo merge on index) the resulting DF directly with your value column.
如果您返回（拆分）位置的系列，您可以将结果 DF 直接与您的值列合并（join以合并索引）。
addr = ['city', 'state', 'region', 'country']
df[['value']].join(df.location.apply(lambda loc: Series(loc, index=addr)))
value city state region country
0 100 Richmond Virginia NaN USA
1 200 New York City New York NaN USA
回答by Martin Alley
I haven't timed this, but I would suggest this option:
我没有计时，但我建议这个选项：
df.loc[:,'city']=df.location.map(lambda x:x[0])
df.loc[:,'state']=df.location.map(lambda x:x[1])
df.loc[:,'regions']=df.location.map(lambda x:x[2])
df.loc[:,'country']=df.location.map(lambda x:x[3])
I'm guessing avoiding explicit for loop might lend itself to a SIMD instruction (certainly numpy looks for that, but perhaps not other libraries)
我猜避免显式 for 循环可能会适合 SIMD 指令（当然 numpy 会寻找它，但可能不是其他库）
