Python保留行（如果YYYY存在），否则删除行

我有一个数据框，它有一个日期列，我想从日期列中删除那些没有YYYY（例如，2018，它可以是任何年份）格式的行。 我曾将apply方法用于正则表达式，但不起作用

df[df.Date.apply(lambda x: re.findall(r'[0-9]{4}', x))]

“日期”列可以具有以下值：

12/3/2018
March 12, 2018
stackoverflow
Mar 12, 2018
no date text
3/12/2018

所以这里的输出应该是

12/3/2018
March 12, 2018
Mar 12, 2018
3/12/2018

---

这是一种方法。使用带有

errors=“胁迫”

Ex:

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018

输出：

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018

或者，如果要维护原始数据

输出：

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018

---

这是一种方法。使用带有

errors=“胁迫”

Ex:

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018

输出：

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018

或者，如果要维护原始数据

输出：

import pandas as pd
df = pd.DataFrame({"Col1": ['12/3/2018', 'March 12, 2018', 'stackoverflow', 'Mar 12, 2018', 'no date text', '3/12/2018']})
df["Col1"] = pd.to_datetime(df["Col1"], errors="coerce")
df = df[df["Col1"].notnull()]
print(df)

        Col1
0 2018-12-03
1 2018-03-12
3 2018-03-12
5 2018-03-12

             Col1
0       12/3/2018
1  March 12, 2018
3    Mar 12, 2018
5       3/12/2018