Question

我有2个要基于列name合并的数据帧。一个df中的name列具有缩写版本，而另一个df中的name列具有全名。更改列名以相互匹配的最有效方法是什么？

df1[names] = ["Man Utd", "Man City", "Chelsea", "Liverpool", "Spurs", "Arsenal"]
df2[names] = ["Manchester United", "Manchester City", "Chelsea FC", "Liverpool FC", "Tottenham Hotspurs", "Arsenal FC"]

Answer 1

您可以使用dict(zip())

如下创建字典

df1['names']  = ["Man Utd", "Man City", "Chelsea", "Liverpool", "Spurs", "Arsenal"]

df2['names']  = ["Manchester United", "Manchester City", "Chelsea FC", "Liverpool FC", "Tottenham Hotspurs", "Arsenal FC"]
d=dict(zip(df1['names'],df2['names'])) #created a mapping dictionary
print(d)

{'Man Utd': 'Manchester United',
'Man City': 'Manchester City',
 'Chelsea': 'Chelsea FC',
 'Liverpool': 'Liverpool FC',
 'Spurs': 'Tottenham Hotspurs',
 'Arsenal': 'Arsenal FC'}

然后通过

更改df1[names]

df1[names]=df1[names].map(d)

在此之后，您可以执行合并，因为列名现在相同。

Answer 2

您可以实现此目标的唯一方法是维持引用it顺序以匹配两个名称列

df1 = pd.DataFrame()

referential = {
    "Man Utd": "Manchester United",
    "Man City": "Manchester City",
    "Chelsea": "Chelsea FC",
    "Liverpool": "Liverpool FC",
    "Spurs": "Tottenham Hotspurs",
    "Arsenal": "Arsenal FC"
}

df1['names'] = ["Man Utd", "Man City", "Chelsea", "Liverpool", "Spurs", "Arsenal"]
df1['names'] = df1['names'].map(referential)
print(df1)

Answer 3

构造字典然后馈入pd.Series.map是一种方法。但是，坚持使用熊猫，您还可以直接使用pd.Series.replace：

lst1 = ["Man Utd", "Man City", "Chelsea", "Liverpool", "Spurs", "Arsenal"]
lst2 = ["Manchester United", "Manchester City", "Chelsea FC", "Liverpool FC",
        "Tottenham Hotspurs", "Arsenal FC"]

# define input dictionary
df = pd.DataFrame({'names': lst1})    

# replace values in lst1 by lst2, by index
df['names'] = df['names'].replace(lst1, lst2)

print(df)

                names
0   Manchester United
1     Manchester City
2          Chelsea FC
3        Liverpool FC
4  Tottenham Hotspurs
5          Arsenal FC

更改一个df中的列值以匹配不同df中的列值？

3 个答案: