如何将跨列的不同二进制值替换为 1/0

Question

我有一个包含多个二进制值的数据集。

df = pd.DataFrame({"a": ["y", "n"], "b": ["t", "f"], 
                   "c": ["known", "unknown"], "d": ['found', 'not found']})

我想将所有二进制列替换为 1/0，同时不影响其他数字列。有没有使用一两行的简单解决方案？数据集包含 500 多列，很难一一检查和替换。谢谢

Answer 1

可以将 pd.get_dummies 与 drop_first=True 归功于@piRSquared

pd.get_dummies(df, drop_first=True)

#   a_y  b_t  c_unknown  d_not found
#0    1    1          0            0
#1    0    0          1            1

如果只需要先对二进制对象列子集执行此操作。

df = pd.DataFrame({'a': ['y', 'n', 'c'], 
                   'b': ['t', 'f', 't'], 
                   'c': ['known', 'unknown', 'known'],
                   'd': ['found', 'not found', 'found'],
                   'e': [1, 2, 2]})

pd.get_dummies(df.loc[:, df.agg('nunique') == 2].select_dtypes(include='object'), 
               drop_first=True)

#   b_t  c_unknown  d_not found
#0    1          0            0
#1    0          1            1
#2    1          0            0

如果跨列的二进制响应数量较少，请考虑创建字典并映射值：

d = {'y': 1, 'n': 0,
     't': 1, 'f': 0,
     'known': 1, 'unknown': 0,
     'found': 1, 'not found': 0}

s = (df.agg('nunique') == 2) & (df.dtypes == 'object')
for col in s[s].index:
    df[col] = df[col].map(d)

#   a  b  c  d  e
#0  y  1  1  1  1
#1  n  0  0  0  2
#2  c  1  1  1  2
#   |
#  `a` not mapped because trinary

如何将跨列的不同二进制值替换为 1/0

How to replace different binary values across columns into 1/0

python

pandas

binary

multiple-conditions