read_table 在 pandas 中，如何将文本输入到数据框
read_table in pandas, how to get input from text to a dataframe
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 pandas 每个 [edit] 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，如果可以怎么办？我确实将所有内容都导入了数据框，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] 行之前的所有值就是它们之间的行的状态
也许pandas可以做到，但你可以轻松做到。
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)
如果您从文件中读取，则
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。
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
这里是代码正在做什么的细目。将文本文件加载到 DataFrame 后，使用 str.contains 来识别状态行。使用 cumsum 对 True/False 个值求和，其中 True 视为 1，False 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 str.replace:
从 states 中删除 [edit] 并从 towns 中删除第一个括号后的所有内容
df['state'] = df['state'].str.replace(r'\[edit\]', '')
df['town'] = df['town'].str.replace(r' \(.+$', '')
删除 town 实际上是状态的行：
df = df.loc[~df['is_state']]
最后，只保留所需的列：
df = df[['state','town']]
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 pandas 每个 [edit] 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，如果可以怎么办？我确实将所有内容都导入了数据框，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] 行之前的所有值就是它们之间的行的状态
也许pandas可以做到，但你可以轻松做到。
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)
如果您从文件中读取，则
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。
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
这里是代码正在做什么的细目。将文本文件加载到 DataFrame 后，使用 str.contains 来识别状态行。使用 cumsum 对 True/False 个值求和，其中 True 视为 1，False 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 str.replace:
states 中删除 [edit] 并从 towns 中删除第一个括号后的所有内容
df['state'] = df['state'].str.replace(r'\[edit\]', '')
df['town'] = df['town'].str.replace(r' \(.+$', '')
删除 town 实际上是状态的行：
df = df.loc[~df['is_state']]
最后，只保留所需的列：
df = df[['state','town']]