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
这与我之前的问题类似:Make a dataframe with grouped questions from three columns
我试着用 stack/pivot 查看以前的问题，但无法弄清楚。实际的数据框有 20 多个问题和李克特量表，从非常同意、同意、中立、不同意、非常不同意。
最佳答案
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
将其提升到另一个层次
我们可以使用 numpy.bincount 来加快速度。但是我们要注意维度
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
%%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
关于python - 统计 pandas 中多列问题的李克特量表结果数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44400661/