pandas 如何使用熊猫计算数据框中每个列表中的元素？
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How to count element in each list in the data frame with pandas?
提问by
Given such a data frame df:
给定这样的数据框df：
0 1
1 [12]
1 [13]
2 [11,12]
1 [10,0,1]
....
I'd like to count a certain value, for instance, '12'in each list of df. So i tried:
我想计算某个值，例如，'12'在每个df. 所以我试过：
df.apply(list.count('12'))
but got error: TypeError: descriptor 'count' requires a 'list' object but received a 'str'. But they are exactly listsin df[1]! How can I correct it? Thanks!
但得到错误：TypeError: descriptor 'count' requires a 'list' object but received a 'str'。但他们正好lists在df[1]！我该如何纠正？谢谢！
采纳答案by jezrael
I think you can try first select column as Series by ixand then applyfunction x.count(12):
我认为您可以尝试先选择列作为系列ix，然后apply运行x.count(12)：
import pandas as pd
d = { 0:pd.Series([1,1,2,1]),
1:pd.Series([[12], [13], [11,12 ],[10,0,1]])}
df = pd.DataFrame(d)
print df
0 1
0 1 [12]
1 1 [13]
2 2 [11, 12]
3 1 [10, 0, 1]
print df.ix[:, 1]
0 [12]
1 [13]
2 [11, 12]
3 [10, 0, 1]
Name: 1, dtype: object
print df.ix[:, 1].apply(lambda x: x.count(12))
0 1
1 0
2 1
3 0
Name: 1, dtype: int64
Or use ilocfor selecting:
或iloc用于选择：
print df.iloc[:, 1].apply(lambda x: x.count(12))
0 1
1 0
2 1
3 0
Name: 1, dtype: int64
EDIT:
编辑：
I think column 1contains NaN.
我认为列1包含NaN.
You can use:
您可以使用：
print df
0 1
0 1 NaN
1 1 [13]
2 2 [11, 12]
3 1 [10, 0, 1]
print df.ix[:, 1].notnull()
0 False
1 True
2 True
3 True
Name: 1, dtype: bool
print df.ix[df.ix[:, 1].notnull(), 1].apply(lambda x: x.count(12))
1 0
2 1
3 0
Name: 1, dtype: int64
EDIT2:
编辑2：
If you want filter by index (e.g. 0:2) and by NaN in column 1:
如果您想按索引（例如0:2）和列中的 NaN过滤1：
print df
0 1
0 1 NaN
1 1 [13]
2 2 [11, 12]
3 1 [10, 0, 1]
#filter df by index - only 0 to 2
print df.ix[0:2, 1]
0 NaN
1 [13]
2 [11, 12]
Name: 1, dtype: object
#boolean series, where is not nul filtered df
print df.ix[0:2, 1].notnull()
0 False
1 True
2 True
Name: 1, dtype: bool
#get column 1: first is filtered to 0:2 index and then if is not null
print df.ix[0:2, 1][df.ix[0:2, 1].notnull()]
1 [13]
2 [11, 12]
Name: 1, dtype: object
#same as above, but more nice
df1 = df.ix[0:2, 1]
print df1
0 NaN
1 [13]
2 [11, 12]
Name: 1, dtype: object
print df1[df1.notnull()]
1 [13]
2 [11, 12]
Name: 1, dtype: object
#apply count
print df1[df1.notnull()].apply(lambda x: x.count(12))
1 0
2 1
Name: 1, dtype: int64
回答by Romain
The counthas to be applied on the column.
在count对列施加。
# Test data
df = pd.DataFrame({1: [[1], [12], [13], [11,12], [10,0,1]]})
df[1].apply(lambda x: x.count(12))
0 0
1 1
2 0
3 1
4 0
Name: 1, dtype: int64
A modification to handle the case when some values are not stored in a list
处理某些值未存储在列表中的情况的修改
# An example with values not stored in list
df = pd.DataFrame({1: [12, [12], [13], [11,12], [10,0,1], 1]})
_check = 12
df[1].apply(lambda l: l.count(_check) if (type(l) is list) else int(l == _check))
0 1
1 1
2 0
3 1
4 0
5 0
Name: 1, dtype: int64
回答by Alexander
You can use a conditional generator expression:
您可以使用条件生成器表达式：
df = df = pd.DataFrame({0: [1, 1, 2, 1, 1, 2], 1: [np.nan, [13], [11, 12], [10, 0, 1], [12], [np.nan, 12]]})
target = 12
>>> sum(sub_list.count(target)
for sub_list in df.iloc[:, 1]
if not np.isnan(sub_list).all())
3
This is like the following conditional list comprehension:
这类似于以下条件列表推导式：
>>> [sub_list.count(12) for sub_list in df.iloc[:, 1] if not np.isnan(sub_list).all()]
[0, 1, 0, 1, 1]
The difference is that the former lazily evaluates each item in the list instead of first generating the entire list, so it is generally more efficient.
区别在于前者懒惰地评估列表中的每个项目，而不是首先生成整个列表，因此通常效率更高。
