Python:使用其他列为Pandas as列表中的新列赋值
我有以下数据框:Python:使用其他列为Pandas as列表中的新列赋值,python,pandas,Python,Pandas,我有以下数据框: Name1 Name2 Score1 Score2 Bruce Jacob 3 4 Aida Stephan 0 1 我想在数据框“list_score”中创建一个新列,它是分数1和2的列表 预期结果: Name1 Name2 Score1 Score2 list_score Bruce Jacob 3 4 [3,4] Aida Stephan 0
Name1 Name2 Score1 Score2
Bruce Jacob 3 4
Aida Stephan 0 1
我想在数据框“list_score”中创建一个新列,它是分数1和2的列表
预期结果:
Name1 Name2 Score1 Score2 list_score
Bruce Jacob 3 4 [3,4]
Aida Stephan 0 1 [0,1]
使用
zip
将元组转换为列表:
df['list_score'] = [list(x) for x in zip(df['Score1'], df['Score2'])]
或:
性能:
df = pd.concat([df] * 1000, ignore_index=True)
In [105]: %timeit df['list_score'] = [list(x) for x in zip(df['Score1'], df['Score2'])]
851 µs ± 36.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [106]: %timeit df['list_score'] = list(map(list, zip(df['Score1'], df['Score2'])))
745 µs ± 35.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [107]: %timeit df['list_score'] = df[['Score1', 'Score2']].apply(tuple, axis=1).apply(list)
35.5 ms ± 295 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [108]: %timeit df['list_score'] = df[['Score1', 'Score2']].values.tolist()
949 µs ± 105 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
这是用于生成上述内容的设置:
一种方法是使用
pd.DataFrame.apply
转换为tuple
,然后再转换为list
。如果tuple
足够,则可以省略第二部分
df['list_score'] = df[['Score1', 'Score2']].apply(tuple, axis=1).apply(list)
print(df)
Name1 Name2 Score1 Score2 list_score
0 Bruce Jacob 3 4 [3, 4]
1 Aida Stephan 0 1 [0, 1]
你是最棒的
def list_comp(df):
df['list_score'] = [list(x) for x in zip(df['Score1'], df['Score2'])]
return df
def map_list(df):
df['list_score'] = list(map(list, zip(df['Score1'], df['Score2'])))
return df
def apply(df):
df['list_score'] = df[['Score1', 'Score2']].apply(tuple, axis=1).apply(list)
return df
def values(df):
df['list_score'] = df[['Score1', 'Score2']].values.tolist()
return df
def make_df(n):
df = pd.DataFrame(np.random.randint(10, size=(n, 2)), columns=['Score1','Score2'])
return df
perfplot.show(
setup=make_df,
kernels=[list_comp, map_list, apply, values],
n_range=[2**k for k in range(2, 15)],
logx=True,
logy=True,
equality_check=False, # rows may appear in different order
xlabel='len(df)')
df['list_score'] = df[['Score1', 'Score2']].apply(tuple, axis=1).apply(list)
print(df)
Name1 Name2 Score1 Score2 list_score
0 Bruce Jacob 3 4 [3, 4]
1 Aida Stephan 0 1 [0, 1]
df['list_score'] = df[['score1', 'score2']].values.tolist()