pandas 比较数据框中的两列值
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compare two columns value in dataframe
提问by Kun OuYang
I have a csv data frame like below, I'd like to compare two column value and generate third column, if value is same will return True, not same return False, how to compare with pandas python?
我有一个如下所示的csv数据框，我想比较两列值并生成第三列，如果值相同将返回True，返回不相同False，如何与pandas python进行比较？
one two
1 a
2 b
3 a
4 b
5 5
6 6
7 7
8 8
9 9
10 10
回答by jezrael
You need if values are mixed (stringand int):
如果值混合（string和int），您需要：
df['three'] = df.one == df.two
But need to_numericif values are not mixed - dtypeof first column is intand second is objectwhat is obviously stringand in column oneare not NaNvalues, because to_numericwith parameter errors='coerce'return NaNfor non numeric values:
但是需要to_numeric如果值不混合 -dtype第一列是int，第二列object显然是什么string，列one中不是NaN值，因为to_numeric参数errors='coerce'返回NaN非数值：
print (pd.to_numeric(df.two, errors='coerce'))
0 NaN
1 NaN
2 NaN
3 NaN
4 5.0
5 6.0
6 7.0
7 8.0
8 9.0
9 10.0
Name: two, dtype: float64
df['three'] = df.one == pd.to_numeric(df.two, errors='coerce')
print (df)
one two three
0 1 a False
1 2 b False
2 3 a False
3 4 b False
4 5 5 True
5 6 6 True
6 7 7 True
7 8 8 True
8 9 9 True
9 10 10 True
Faster solution with Series.eq:
更快的解决方案Series.eq：
df['three'] = df.one.eq(pd.to_numeric(df.two, errors='coerce'))
print (df)
one two three
0 1 a False
1 2 b False
2 3 a False
3 4 b False
4 5 5 True
5 6 6 True
6 7 7 True
7 8 8 True
8 9 9 True
9 10 10 True
