计算pandas中多列问题的likert量表结果数

Posted 2023-03-11

【中文标题】计算pandas中多列问题的likert量表结果数

【英文标题】：Count the number of likert scale results from multiple column questions in pandas

【发布时间】：2017-11-08 02:22:59

【问题描述】：

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下面的？

              Agree         Disagree         Strongly Disagree
 Question1    2               1                  0

 Question2    2               1                  0

 Question3    2               1                  0
 Question4    1               1                  1

这和我之前的问题类似：Make a dataframe with grouped questions from three columns

我尝试使用 stack/pivot 查看以前的问题，但无法弄清楚。实际的数据框有 20 多个问题和一个李克特量表，从非常同意、同意、中立、不同意、非常不同意。

【参考方案1】：

您可以使用pd.Series.value_counts 遍历列。如果您使用 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

一点后处理：

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

【参考方案2】：

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

【参考方案3】：

与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

---

将其提升到另一个层次我们可以使用 numpy.bincount 来加快速度。但是我们要注意尺寸

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

---

速度差异

%%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

【讨论】：

谢谢！这非常有效，我上一个问题中的“额外积分”部分帮助我对列进行了排序。