我有一個這樣的數(shù)據(jù)框dfcol1 col2 col3 A black berry black B green apple green C red wine red我想從 col2 值中減去 col3 值，結(jié)果看起來像df1col1 col2 col3 A berry black B apple green C wine red如何使用熊貓以有效的方式做到這一點
1 回答
揚帆大魚
TA貢獻(xiàn)1799條經(jīng)驗 獲得超9個贊
list comprehension與replace和 一起使用split：
df['col2'] = [a.replace(b, '').strip() for a, b in zip(df['col2'], df['col3'])]
print (df)
col1 col2 col3
0 A berry black
1 B apple green
2 C wine red
如果順序不重要，則將拆分的值轉(zhuǎn)換為集合并減去：
df['col2'] = [' '.join(set(a.split())-set([b])) for a, b in zip(df['col2'], df['col3'])]
print (df)
col1 col2 col3
0 A berry black
1 B apple green
2 C wine red
或者使用帶有if條件和的生成器join：
df['col2'] = [' '.join(c for c in a.split() if c != b) for a, b in zip(df['col2'], df['col3'])]
性能:
這是用于生成上面的perfplot的設(shè)置：
def calculation(val):
return val[0].replace(val[1],'').strip()
def regex(df):
df.col2=df.col2.replace(regex=r'(?i)'+ df.col3,value="")
return df
def lambda_f(df):
df["col2"] = df.apply(lambda x: x["col2"].replace(x["col3"], "").strip(), axis=1)
return df
def apply(df):
df['col2'] = df[['col2','col3']].apply(calculation, axis=1)
return df
def list_comp1(df):
df['col2'] = [a.replace(b, '').strip() for a, b in zip(df['col2'], df['col3'])]
return df
def list_comp2(df):
df['col2'] = [' '.join(set(a.split())-set([b])) for a, b in zip(df['col2'], df['col3'])]
return df
def list_comp3(df):
df['col2'] = [' '.join(c for c in a.split() if c != b) for a, b in zip(df['col2'], df['col3'])]
return df
def make_df(n):
d = {'col1': {0: 'A', 1: 'B', 2: 'C'}, 'col2': {0: 'black berry', 1: 'green apple', 2: 'red wine'}, 'col3': {0: 'black', 1: 'green', 2: 'red'}}
df = pd.DataFrame(d)
df = pd.concat([df] * n * 100, ignore_index=True)
return df
perfplot.show(
setup=make_df,
kernels=[regex, lambda_f, apply, list_comp1,list_comp2,list_comp3],
n_range=[2**k for k in range(2, 10)],
logx=True,
logy=True,
equality_check=False, # rows may appear in different order
xlabel='len(df)')
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