python - 使用 Dataframe 中的现有列分配列值时出现问题

Question

我正在尝试在数据框中创建一个新列，该列根据另一列中的值分配值。我正在使用的代码分配值，但不是我想要的。我不确定我错过了什么。

代码示例如下:

#define track styles
short = [4,6,8,9,11,20,24,28,30,33,35]
inter = [2,3,7,12,13,17,19,25,27,32,34,36]
long = [5,14,15,21,23,26]
plate = [1,10,18,31]
road = [16,22,29]

#input driver and stat info    
driver1 = input('Choose driver: ')

#read driver data to dataframe
df = pd.read_csv(driver1 + '_2018.csv')

#add track type
df['Type'] = ''

for i in range(len(df)):
    if df['Race'][i] in short:
        df['Type'][i] = 'short'
    elif df['Race'][i] in inter:
        df['Type'] = 'intermediate'
    elif df['Race'][i] in long:
        df['Type'] = 'long'
    elif df['Race'][i] in plate:
        df['Type'] = 'plate'
    else:
        df['Type'] = 'road'

print(df.head())

我得到以下输出:

C:\EclipseWorkspace\csse120\Personal\NASCAR_Projects\Other\driver_review.py:45: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  df['Type'][i] = 'short'
   Race  Start  Mid Race      ...       Total Laps  DRIVER RATING          Type
0     1      5        23      ...              207          105.2  intermediate
1     2     16         7      ...              325           94.2  intermediate
2     3     10         2      ...              267          106.1  intermediate
3     4      5        11      ...              311           80.0  intermediate
4     5      6         3      ...              200          113.0  intermediate

[5 rows x 20 columns]

请注意，“类型”列返回所有“中间”，而它应该包括 ['plate'、'intermediate'、'intermediate'、'short'、'long']。

Answer 1

使用map和 dictionary - 首先通过键中的新名称创建字典并在值中列出，然后在字典理解中将彼此交换为平面字典:

d = {'short':short, 
     'intermediate':inter,
     'long':long,
     'plate':plate,
     'road':road}

d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}
df['Type'] = df['Race'].map(d1)
print (df)
   Race  Start  Mid Race  Total Laps  DRIVER RATING          Type
0     1      5        23         207          105.2         plate
1     2     16         7         325           94.2  intermediate
2     3     10         2         267          106.1  intermediate
3     4      5        11         311           80.0         short
4     5      6         3         200          113.0          long

如果想要将前 4 个类别中不匹配的所有值设置为 road 从第一个字典中删除 road 并添加 fillna替换所有不匹配的值:

d = {'short':short, 
     'intermediate':inter,
     'long':long,
     'plate':plate}

d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}
df['Type'] = df['Race'].map(d1).fillna('road')

详细信息:

print (d1)

{
    4: 'short', 6: 'short',
    8: 'short', 9: 'short',
    11: 'short',    20: 'short',
    24: 'short',    28: 'short',
    30: 'short',    33: 'short',
    35: 'short',    2: 'intermediate',
    3: 'intermediate',  7: 'intermediate',
    12: 'intermediate', 13: 'intermediate',
    17: 'intermediate', 19: 'intermediate',
    25: 'intermediate', 27: 'intermediate',
    32: 'intermediate', 34: 'intermediate',
    36: 'intermediate', 5: 'long',
    14: 'long', 15: 'long',
    21: 'long', 23: 'long',
    26: 'long', 1: 'plate',
    10: 'plate',    18: 'plate',
    31: 'plate',    16: 'road',
    22: 'road', 29: 'road'
}