Pandas：获取所有具有常量值的列
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Pandas: Get all columns that have constant value
提问by tbienias
I want to get the names of the columns which have same values across all rows for each column.
我想获取在每列的所有行中具有相同值的列的名称。
My data:
我的数据：
A B C D
0 1 hi 2 a
1 3 hi 2 b
2 4 hi 2 c
Desired output:
期望的输出：
['B', 'C']
Code:
代码：
import pandas as pd
d = {'A': [1,3,4], 'B': ['hi','hi','hi'], 'C': [2,2,2], 'D': ['a','b','c']}
df = pd.DataFrame(data=d)
I've been playing around with df.columnsand .any(), but can't figure out how to do this.
我一直在玩df.columns和.any()，但不知道如何做到这一点。
回答by smci
Use the pandas not-so-well-known builtin nunique():
使用不太知名的内置Pandasnunique()：
df.columns[df.nunique() <= 1]
Index(['B', 'C'], dtype='object')
Notes:
笔记：
- Use
dropna=Falseoption if you want na's counted as a separate value - It's the cleanest code, but not the fastest
dropna=False如果您希望将 na 计为单独的值，请使用选项- 这是最干净的代码，但不是最快的
回答by jezrael
Solution 1:
解决方案1：
c = [c for c in df.columns if len(set(df[c])) == 1]
print (c)
['B', 'C']
Solution 2:
解决方案2：
c = df.columns[df.eq(df.iloc[0]).all()].tolist()
print (c)
['B', 'C']
Explanation for Solution 2:
解决方案 2 的说明：
First compare all rows to the first row with DataFrame.eq...
首先将所有行与第一行进行比较DataFrame.eq...
print (df.eq(df.iloc[0]))
A B C D
0 True True True True
1 False True True False
2 False True True False
... then check each column is all Trues with DataFrame.all...
...然后检查每一列都是Trues 与DataFrame.all...
print (df.eq(df.iloc[0]).all())
A False
B True
C True
D False
dtype: bool
... finally filter columns' names for which result is True:
... 最后过滤结果为 True 的列名称：
print (df.columns[df.eq(df.iloc[0]).all()])
Index(['B', 'C'], dtype='object')
Timings:
时间：
np.random.seed(100)
df = pd.DataFrame(np.random.randint(10, size=(1000,100)))
df[np.random.randint(100, size=20)] = 100
print (df)
# Solution 1 (second-fastest):
In [243]: %timeit ([c for c in df.columns if len(set(df[c])) == 1])
3.59 ms ± 43.8 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# Solution 2 (fastest):
In [244]: %timeit df.columns[df.eq(df.iloc[0]).all()].tolist()
1.62 ms ± 13.3 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
#Mohamed Thasin ah solution
In [245]: %timeit ([col for col in df.columns if len(df[col].unique())==1])
6.8 ms ± 352 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#jpp solution
In [246]: %%timeit
...: vals = df.apply(set, axis=0)
...: res = vals[vals.map(len) == 1].index
...:
5.59 ms ± 64.7 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#smci solution 1
In [275]: %timeit df.columns[ df.nunique()==1 ]
11 ms ± 105 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#smci solution 2
In [276]: %timeit [col for col in df.columns if not df[col].is_unique]
9.25 ms ± 80 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#smci solution 3
In [277]: %timeit df.columns[ df.apply(lambda col: not col.is_unique) ]
11.1 ms ± 511 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
回答by Mohamed Thasin ah
try this,
尝试这个，
print [col for col in df.columns if len(df[col].unique())==1]
Output:
输出：
['B', 'C']
回答by jpp
You can use setand apply a filter on a series:
您可以set在系列上使用和应用过滤器：
vals = df.apply(set, axis=0)
res = vals[vals.map(len) == 1].index
print(res)
Index(['B', 'C'], dtype='object')
Use res.tolist()if having a list output is important.
使用res.tolist()如果有一个列表输出是很重要的。
