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Use a values of list to filter Pandas dataframe

With python Pandas, I'm trying to filter out the data that contains the specified value in the array, I try to use python in to filter value, but it's not working, I want to know if there is a way to achieve such a function without looping
import pandas as pd
df = pd.DataFrame({'A' : [1,2,3,4], 'B' : [[1, 2, 3], [2, 3], [3], [1, 2, 3]]})
df = 1 in df['custom_test_type']
A B
0 1 [1, 2, 3]
1 2 [2, 3]
2 3 [3]
3 4 [1, 2, 3]
I'm try to filter 1 in row B, so expected output will be:
A B
0 1 [1, 2, 3]
3 4 [1, 2, 3]
but the output always be True
due to my limited ability, Any help or explanation is welcome! Thank you.

You need to use a loop/list comprehension:
out = df[[1 in l for l in df['B']]]
A pandas version would be more verbose and less efficient:
out = df[df['B'].explode().eq(1).groupby(level=0).any()]
Output:
A B
0 1 [1, 2, 3]
3 4 [1, 2, 3]

remove elements from list based on index in pandas Dataframe

How to remove elements from list based on index range in pandas Dataframe.
suppose DataFrame is like
df:
values size
0 [1,2,3,4,5,6,7] 2 #delete first 2 elements from list
1 [1,2,3,4] 3 #delete first 3 elements from list
2 [9,8,7,6,5,4,3] 5 #delete first 5 elements from list
Expected Output is
df:
values size
0 [3,4,5,6,7] 2
1 [4] 3
2 [4,3] 5

Use list comprehension with indexing:
df['values'] = [i[j:] for i, j in zip(df['values'], df['size'])]
print (df)
values size
0 [3, 4, 5, 6, 7] 2
1 [4] 3
2 [4, 3] 5

Using df.apply
import pandas as pd
df = pd.DataFrame({"values": [[1,2,3,4,5,6,7], [1,2,3,4], [9,8,7,6,5,4,3]], "size": [2, 3, 5]})
df["values"] = df.apply(lambda x: x["values"][x['size']:], axis=1)
print(df)
Output:
size values
0 2 [3, 4, 5, 6, 7]
1 3 [4]
2 5 [4, 3]

Using map in base Python, you could do
dat['values'] = pd.Series(map(lambda x, y : x[y:], dat['values'], dat['size']))
which returns
dat
Out[34]:
values size
0 [3, 4, 5, 6, 7] 2
1 [4] 3
2 [4, 3] 5

Python - pick a value from a list basing on another list

I've got a dataframe. In column A there is a list of integers, in column B - an integer. I want to pick n-th value of the column A list, where n is a number from column B. So if in columns A there is [1,5,6,3,4] and in column B: 2, I want to get '6'.
I tried this:
result = [y[x] for y in df['A'] for x in df['B']
But it doesn't work. Please help.

Use zip with list comprehension:
df['new'] = [y[x] for x, y in zip(df['B'], df['A'])]
print (df)
A B new
0 [1, 2, 3, 4, 5] 1 2
1 [1, 2, 3, 4] 2 3

You can go for apply i.e
df = pd.DataFrame({'A':[[1,2,3,4,5],[1,2,3,4]],'B':[1,2]})
A B
0 [1, 2, 3, 4, 5] 1
1 [1, 2, 3, 4] 2
# df.apply(lambda x : np.array(x['A'])[x['B']],1)
# You dont need np.array here, use it when the column B is also a list.
df.apply(lambda x : x['A'][x['B']],1) # Thanks #Zero
0 2
1 3
dtype: int64

Find all duplicate rows in a pandas dataframe

I would like to be able to get the indices of all the instances of a duplicated row in a dataset without knowing the name and number of columns beforehand. So assume I have this:
col
1 | 1
2 | 2
3 | 1
4 | 1
5 | 2
I'd like to be able to get [1, 3, 4] and [2, 5]. Is there any way to achieve this? It sounds really simple, but since I don't know the columns beforehand I can't do something like df[col == x...].

First filter all duplicated rows and then groupby with apply or convert index to_series:
df = df[df.col.duplicated(keep=False)]
a = df.groupby('col').apply(lambda x: list(x.index))
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object
a = df.index.to_series().groupby(df.col).apply(list)
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object
And if need nested lists:
L = df.groupby('col').apply(lambda x: list(x.index)).tolist()
print (L)
[[1, 3, 4], [2, 5]]
If need use only first column is possible selected by position with iloc:
a = df[df.iloc[:,0].duplicated(keep=False)]
.groupby(df.iloc[:,0]).apply(lambda x: list(x.index))
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object

python pandas filtering involving lists

I am currently using Pandas in python 2.7. My dataframe looks similar to this:
>>> df
0
1 [1, 2]
2 [2, 3]
3 [4, 5]
Is it possible to filter rows by values in column 1? For example, if my filter value is 2, the filter should return a dataframe containing the first two rows.
I have tried a couple of ways already. The best thing I can think of is to do a list comprehension that returns the index of rows in which the value exist. Then, I could filter the dataframe with the list of indices. But, this would be very slow if I want to filter multiple times with different values. Ideally, I would like something that uses the build in Pandas functions in order to speed up the process.

You can use boolean indexing:
import pandas as pd
df = pd.DataFrame({'0':[[1, 2],[2, 3], [4, 5]]})
print (df)
0
0 [1, 2]
1 [2, 3]
2 [4, 5]
print (df['0'].apply(lambda x: 2 in x))
0 True
1 True
2 False
Name: 0, dtype: bool
print (df[df['0'].apply(lambda x: 2 in x)])
0
0 [1, 2]
1 [2, 3]

You can also use boolean indexing with a list comprehension:
>>> df[[2 in row for row in df['0']]]
0
0 [1, 2]
1 [2, 3]