pandas 如何根据pandas中其他列的值计算新列-python
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how to compute a new column based on the values of other columns in pandas - python
提问by HappyPy
Let's say my data frame contains these data:
假设我的数据框包含这些数据：
>>> df = pd.DataFrame({'a':['l1','l2','l1','l2','l1','l2'],
'b':['1','2','2','1','2','2']})
>>> df
a b
0 l1 1
1 l2 2
2 l1 2
3 l2 1
4 l1 2
5 l2 2
l1should correspond to 1whereas l2should correspond to 2.
I'd like to create a new column 'c' such that, for each row, c = 1if a = l1and b = 1(or a = l2and b = 2). If a = l1and b = 2(or a = l2and b = 1) then c = 0.
l1应该对应1而l2应该对应2。我想创建一个新列 ' c'，对于每一行，c = 1如果a = l1和b = 1（或a = l2和b = 2）。如果a = l1和b = 2（或a = l2和b = 1）那么c = 0。
The resulting data frame should look like this:
生成的数据框应如下所示：
a b c
0 l1 1 1
1 l2 2 1
2 l1 2 0
3 l2 1 0
4 l1 2 0
5 l2 2 1
My data frame is very large so I'm really looking for the most efficient way to do this using pandas.
我的数据框非常大，所以我真的在寻找使用 Pandas 执行此操作的最有效方法。
回答by chlunde
df = pd.DataFrame({'a': numpy.random.choice(['l1', 'l2'], 1000000),
'b': numpy.random.choice(['1', '2'], 1000000)})
A fast solution assuming only two distinct values:
假设只有两个不同值的快速解决方案：
%timeit df['c'] = ((df.a == 'l1') == (df.b == '1')).astype(int)
10 loops, best of 3: 178 ms per loop
10 个循环，最好的 3 个：每个循环 178 毫秒
@Viktor Kerkes:
@维克多·凯克斯：
%timeit df['c'] = (df.a.str[-1] == df.b).astype(int)
1 loops, best of 3: 412 ms per loop
1 个循环，最好的 3 个：每个循环 412 毫秒
@user1470788:
@用户1470788：
%timeit df['c'] = (((df['a'] == 'l1')&(df['b']=='1'))|((df['a'] == 'l2')&(df['b']=='2'))).astype(int)
1 loops, best of 3: 363 ms per loop
1 个循环，最好的 3 个：每个循环 363 毫秒
@herrfz
@herrfz
%timeit df['c'] = (df.a.apply(lambda x: x[1:])==df.b).astype(int)
1 loops, best of 3: 387 ms per loop
1 个循环，最好的 3 个：每个循环 387 毫秒
回答by Viktor Kerkez
You can also use the string methods.
您还可以使用字符串方法。
df['c'] = (df.a.str[-1] == df.b).astype(int)
回答by herrfz
df['c'] = (df.a.apply(lambda x: x[1:])==df.b).astype(int)
df['c'] = (df.a.apply(lambda x: x[1:])==df.b).astype(int)
回答by user1470788
You can just use logical operators. I'm not sure why you're using strings of 1 and 2 rather than ints, but here's a solution. The astype at the end converts it from boolean to 0's and 1's.
您可以只使用逻辑运算符。我不确定您为什么使用 1 和 2 的字符串而不是整数，但这是一个解决方案。最后的 astype 将它从布尔值转换为 0 和 1。
df['c'] = (((df['a'] == 'l1')&(df['b']=='1'))|((df['a'] == 'l2')&(df['b']=='2'))).astype(int)
df['c'] = (((df['a'] == 'l1')&(df['b']=='1'))|((df['a'] == 'l2')&(df['b']=='2'))).astype(int)
