英文:
Is it possible to split a pandas column after last integer?
问题
I am trying to split a pandas column into two separate, where the first should contain just the date and the second string. But I don't want to split it after a certain character, like counting where the last integer instead I want to make a code that is applicable in general.
My col looks like this :
| Column A |
|---|
| 01.01.2000John Doe |
| 01.01.2002Jane Doe |
And I want it to look like this:
| Column A | Column B |
|---|---|
| 01.01.2000 | Johne Doe |
| 01.01.2001 | Jane Doe |
英文:
I am trying to split a pandas column into two separate, where the first should contain just the date and the second string. But I don't want to split it after a certain character, like counting where the last integer instead I want to make a code that is applicable in general.
My col looks like this :
| Column A |
|---|
| 01.01.2000John Doe |
| 01.01.2002Jane Doe |
And I want it to look like this:
| Column A | Column B |
|---|---|
| 01.01.2000 | Johne Doe |
| 01.01.2001 | Jane Doe |
df_t['date'] = df_t['date_time'].str[0:19]df_t["name"] = df_t["date_time"].str[19: ]tid = df_t.drop(["date_time"], axis = 1)
This is the way I did it but I need a general way as mentioned above
答案1
得分: 1
You can use str.extract along with regular expressions in your code to extract date and name information from the given data.
import pandas as pd# 示例数据data = {'Column A': ['01.01.2000John Doe', '01.01.2002Jane Doe']}df = pd.DataFrame(data)# 正则表达式模式pattern = r'(?P<Date>\d{2}\.\d{2}\.\d{4})(?P<Name>.*)&'# 将日期和姓名提取到单独的列中df[['Column A', 'Column B']] = df['Column A'].str.extract(pattern)print(df)
Explanation:
pattern变量包含了正则表达式模式。表达式(?P<Date>\d{2}\.\d{2}\.\d{4})用于捕获日期，而(?P<Name>.*)用于捕获姓名。- 使用
?P<>语法对捕获的组进行命名，以便更容易在DataFrame中创建新列。
英文:
You can use str.extract together with regular expressions:
import pandas as pd# Sample datadata = {'Column A': ['01.01.2000John Doe', '01.01.2002Jane Doe']}df = pd.DataFrame(data)# Regular expression patternpattern = r'(?P<Date>\d{2}\.\d{2}\.\d{4})(?P<Name>.*)'# Extracting the date and name into separate columnsdf[['Column A', 'Column B']] = df['Column A'].str.extract(pattern)print(df)
Explanation:
- The pattern variable contains the regular expression pattern. The expression (?P<Date>\d{2}.\d{2}.\d{4}) captures the date, and (?P<Name>.*) captures the name.
- The ?P<> syntax is used to name the captured groups, which makes it easier to create the new columns in the DataFrame.
EDIT
import pandas as pd# Sample datadata = {'1Column A': ['2000-01-01 00:00:00John Doe', '2002-01-01 00:00:00Jane Doe'],'2Column B': ['2000-01-01 00:00:00Alice', '2002-01-01 00:00:00Bob'],'3Column C': ['Some other data', 'Not a date and name'],}df = pd.DataFrame(data)# Regular expression patternpattern = r'(?P<Date>\d{4}-\d{2}-\d{2} \d{2}:\d{2}:\d{2})(?P<Name>.*)'# Iterate through columns and apply the pattern conditionallyfor col in df.columns:if col.startswith("1") or col.startswith("2"):# Extract date and name into separate columns with suffixesdf[[f"{col}_date", f"{col}_name"]] = df[col].str.extract(pattern)# Drop the original columndf.drop(col, axis=1, inplace=True)print(df)
答案2
得分: 1
你可以简单地使用索引：
df['Column A'], df['Column B'] = df['Column A'].str[:10], df['Column A'].str[10:]print(df)# 输出Column A Column B0 01.01.2000 John Doe1 01.01.2002 Jane Doe
如果你想转换成datetime64：
df['Column A'], df['Column B'] = \pd.to_datetime(df['Column A'].str[:10], dayfirst=True), df['Column A'].str[10:]print(df)# 输出Column A Column B0 2000-01-01 John Doe1 2002-01-01 Jane Doe
英文:
You can simply use indexing:
df['Column A'], df['Column B'] = df['Column A'].str[:10], df['Column A'].str[10:]print(df)# OutputColumn A Column B0 01.01.2000 John Doe1 01.01.2002 Jane Doe
If you want to convert as datetime64:
df['Column A'], df['Column B'] = \pd.to_datetime(df['Column A'].str[:10], dayfirst=True), df['Column A'].str[10:]print(df)# OutputColumn A Column B0 2000-01-01 John Doe1 2002-01-01 Jane Doe
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