有什么方法可以根據來自另一列的字串匹配值創建一個新列，而無需復制該列并替換這些值？
到目前為止，這是我正在使用的一種解決方案：
df = pd.DataFrame({
"Task": ["T1", "T1", "T2", "T3","T3"],
"Complexity": [1, 2, 3, 2, 1],
"Status": ["OpenA", "ClosedB", "ClosedC", "OpenC", "OpenA"]})
df['Statno']=df['Status']
df=df.replace({'Statno':{'(?i).*Open.*': '1', '(?i).*Close.*' : '0' }}, regex=True)
df
uj5u.com熱心網友回復：
像那樣？
import pandas as pd
df = pd.DataFrame({
"Task": ["T1", "T1", "T2", "T3","T3"],
"Complexity": [1, 2, 3, 2, 1],
"Status": ["OpenA", "ClosedB", "ClosedC", "OpenC", "OpenA"]})
df['Statno']=df['Status'].replace({'(?i).*Open.*': '1', '(?i).*Close.*' : '0' }, regex=True)
df
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標籤：Python熊猫细绳代替