Question

我有以下数据框：

       Question1        Question2         Question3          Question4
User1  Agree            Agree          Disagree         Strongly Disagree
User2  Disagree         Agree          Agree            Disagree
User3  Agree            Agree          Agree            Agree

有没有办法将上面列出的数据框转换为以下内容？

              Agree         Disagree         Strongly Disagree
 Question1    2               1                  0

 Question2    2               1                  0

 Question3    2               1                  0
 Question4    1               1                  1

这与我之前的问题相似：Make a dataframe with grouped questions from three columns

我尝试使用stack / pivot查看之前的问题，但无法弄明白。实际的数据框架有20多个问题和一个强烈同意，同意，中立，不同意，强烈反对的缩写。

Answer 1

使用pd.get_dummies

pd.get_dummies(df.stack()).groupby(level=1).sum()

           Agree  Disagree  Strongly Disagree
Question1      2         1                  0
Question2      3         0                  0
Question3      2         1                  0
Question4      1         1                  1

将其提升到另一个级别
我们可以使用numpy.bincount来加快速度。但我们必须注意尺寸

v = df.values
f, u = pd.factorize(v.ravel())
n, m = u.size, v.shape[1]
r = np.tile(np.arange(m), n)
b0 = np.bincount(r * n + f)
pad = np.zeros(n * m - b0.size, dtype=int)
b = np.append(b0, pad)

pd.DataFrame(b.reshape(m, n), df.columns, u)

           Agree  Disagree  Strongly Disagree
Question1      2         1                  0
Question2      3         0                  0
Question3      2         1                  0
Question4      1         1                  1

另一个numpy选项

v = df.values
n, m = v.shape
f, u = pd.factorize(v.ravel())

pd.DataFrame(
    np.eye(u.size, dtype=int)[f].reshape(n, m, -1).sum(0),
    df.columns, u
)

           Agree  Disagree  Strongly Disagree
Question1      2         1                  0
Question2      3         0                  0
Question3      2         1                  0
Question4      1         1                  1

速差

%%timeit
v = df.values
f, u = pd.factorize(v.ravel())
n, m = u.size, v.shape[1]
r = np.tile(np.arange(m), n)
b0 = np.bincount(r * n + f)
pad = np.zeros(n * m - b0.size, dtype=int)
b = np.append(b0, pad)

pd.DataFrame(b.reshape(m, n), df.columns, u)
1000 loops, best of 3: 194 µs per loop

%%timeit
v = df.values
n, m = v.shape
f, u = pd.factorize(v.ravel())

pd.DataFrame(
    np.eye(u.size, dtype=int)[f].reshape(n, m, -1).sum(0),
    df.columns, u
)
1000 loops, best of 3: 195 µs per loop

%timeit pd.get_dummies(df.stack()).groupby(level=1).sum()
1000 loops, best of 3: 1.2 ms per loop

Answer 2

您可以使用pd.Series.value_counts迭代列。如果使用apply执行此操作，索引将自动对齐：

df.apply(pd.Series.value_counts)
Out: 
                   Question1  Question2  Question3  Question4
Agree                    2.0        3.0        2.0          1
Disagree                 1.0        NaN        1.0          1
Strongly Disagree        NaN        NaN        NaN          1

稍微后处理：

df.apply(pd.Series.value_counts).fillna(0).astype('int')
Out: 
                   Question1  Question2  Question3  Question4
Agree                      2          3          2          1
Disagree                   1          0          1          1
Strongly Disagree          0          0          0          1

Answer 3

df.apply(lambda x:x.value_counts()).fillna(0).astype(int)
#                   Question1  Question2  Question3  Question4
#Agree                      2          3          2          1
#Disagree                   1          0          1          1
#Strongly Disagree          0          0          0          1

从熊猫中的多列问题中计算出Likert量表结果的数量

3 个答案: