1 
Alabama[edit]
Auburn (Auburn University)[1]
Florence (University of North Alabama)
Jacksonville (Jacksonville State University)[2]
Alaska[edit]
Fairbanks (University of Alaska Fairbanks)[2]
Arizona[edit]
Flagstaff (Northern Arizona University)[6]
Tempe (Arizona State University)
Tucson (University of Arizona)

这是我的文本，我需要创建一个数据框，其中 1 列用于州名称，另一列用于城镇名称，我知道如何删除大学名称。但是我如何告诉熊猫每次[编辑]都是一个新状态。

预期输出数据帧

Alabama Auburn
Alabama Florence 
Alabama Jacksonville
Alaska  Fairbanks 
Arizona Flagstaff
Arizona Tempe
Arizona Tucson

我不确定我是否可以使用 read_table，如果可以的话？我确实将所有内容都导入了数据框中，但州和城市在同一列中。我也尝试了一个列表，但问题仍然是一样的。

我需要一些东西，如果该行中有一个 [edit]，那么它之后和下一个 [edit] 行之前的所有值都是中间行的状态

4

2 回答 2

4

使用 Pandas，您可以执行以下操作：

import pandas as pd
df = pd.read_table('data', sep='\n', header=None, names=['town'])
df['is_state'] = df['town'].str.contains(r'\[edit\]')
df['groupno'] = df['is_state'].cumsum()
df['index'] = df.groupby('groupno').cumcount()
df['state'] = df.groupby('groupno')['town'].transform('first')
df['state'] = df['state'].str.replace(r'\[edit\]', '')
df['town'] = df['town'].str.replace(r' \(.+$', '')
df = df.loc[~df['is_state']]
df = df[['state','town']]

产生

     state          town
1  Alabama        Auburn
2  Alabama      Florence
3  Alabama  Jacksonville
5   Alaska     Fairbanks
7  Arizona     Flagstaff
8  Arizona         Tempe
9  Arizona        Tucson

这是代码正在执行的操作的细分。将文本文件加载到 DataFrame 后，用于str.contains识别哪些行是状态。用于cumsum获取 True/False 值的累积和，其中 True 被视为 1，False 被视为 0。

df = pd.read_table('data', sep='\n', header=None, names=['town'])
df['is_state'] = df['town'].str.contains(r'\[edit\]')
df['groupno'] = df['is_state'].cumsum()
#                                               town is_state  groupno
# 0                                    Alabama[edit]     True        1
# 1                    Auburn (Auburn University)[1]    False        1
# 2           Florence (University of North Alabama)    False        1
# 3  Jacksonville (Jacksonville State University)[2]    False        1
# 4                                     Alaska[edit]     True        2
# 5    Fairbanks (University of Alaska Fairbanks)[2]    False        2
# 6                                    Arizona[edit]     True        3
# 7       Flagstaff (Northern Arizona University)[6]    False        3
# 8                 Tempe (Arizona State University)    False        3
# 9                   Tucson (University of Arizona)    False        3

现在对于每个groupno数字，我们可以为组中的每一行分配一个唯一的整数：

df['index'] = df.groupby('groupno').cumcount()
#                                               town is_state  groupno  index
# 0                                    Alabama[edit]     True        1      0
# 1                    Auburn (Auburn University)[1]    False        1      1
# 2           Florence (University of North Alabama)    False        1      2
# 3  Jacksonville (Jacksonville State University)[2]    False        1      3
# 4                                     Alaska[edit]     True        2      0
# 5    Fairbanks (University of Alaska Fairbanks)[2]    False        2      1
# 6                                    Arizona[edit]     True        3      0
# 7       Flagstaff (Northern Arizona University)[6]    False        3      1
# 8                 Tempe (Arizona State University)    False        3      2
# 9                   Tucson (University of Arizona)    False        3      3

同样对于每个groupno数字，我们可以通过选择每个组中的第一个城镇来找到该州：

df['state'] = df.groupby('groupno')['town'].transform('first')
#                                               town is_state  groupno  index          state
# 0                                    Alabama[edit]     True        1      0  Alabama[edit]
# 1                    Auburn (Auburn University)[1]    False        1      1  Alabama[edit]
# 2           Florence (University of North Alabama)    False        1      2  Alabama[edit]
# 3  Jacksonville (Jacksonville State University)[2]    False        1      3  Alabama[edit]
# 4                                     Alaska[edit]     True        2      0   Alaska[edit]
# 5    Fairbanks (University of Alaska Fairbanks)[2]    False        2      1   Alaska[edit]
# 6                                    Arizona[edit]     True        3      0  Arizona[edit]
# 7       Flagstaff (Northern Arizona University)[6]    False        3      1  Arizona[edit]
# 8                 Tempe (Arizona State University)    False        3      2  Arizona[edit]
# 9                   Tucson (University of Arizona)    False        3      3  Arizona[edit]

我们基本上有了想要的DataFrame；剩下的就是美化结果。我们可以使用 s[edit]从s中删除state第一个括号之后的所有内容：townstr.replace

df['state'] = df['state'].str.replace(r'\[edit\]', '')
df['town'] = df['town'].str.replace(r' \(.+$', '')

删除town实际是状态的行：

df = df.loc[~df['is_state']]

最后，只保留所需的列：

df = df[['state','town']]
于 2016-11-04T01:13:52.797 回答
3

也许pandas可以做到，但您可以轻松做到。

data = '''Alabama[edit]
Auburn (Auburn University)[1]
Florence (University of North Alabama)
Jacksonville (Jacksonville State University)[2]
Alaska[edit]
Fairbanks (University of Alaska Fairbanks)[2]
Arizona[edit]
Flagstaff (Northern Arizona University)[6]
Tempe (Arizona State University)
Tucson (University of Arizona)'''

# ---

result = []

state = None

for line in data.split('\n'):

    if line.endswith('[edit]'):
        # remember new state
        state = line[:-6] # without `[edit]`
    else:
        # add state, city to result
        city, rest = line.split(' ', 1)
        result.append( [state, city] )

# --- display ---

for state, city in result:
    print(state, city)

如果你从文件中读取然后

result = []

state = None

with open('your_file') as f:
    for line in f:
        line = line.strip() # remove '\n'

        if line.endswith('[edit]'):
            # remember new state
            state = line[:-6] # without `[edit]`
        else:
            # add state, city to result
            city, rest = line.split(' ', 1)
            result.append( [state, city] )

# --- display ---

for state, city in result:
    print(state, city)

现在您可以使用result创建DataFrame.

于 2016-11-04T00:44:24.960 回答