value Group something
0 a 1 1
1 b 1 2
2 c 1 4
3 c 2 9
4 b 2 10
5 x 2 5
6 d 2 3
7 e 3 5
8 d 2 10
9 a 3 5
I want to select the last 3 rows of each group(from the above df) like the following but perform the operation using Inplace. I want to ensure that I am keeping only the new df object in memory after assignment. What would be an efficient way of doing it?
df = df.groupby('Group').tail(3)
The result should look like the following:
value Group something
0 a 1 1
1 b 1 2
2 c 1 4
5 x 2 5
6 d 2 3
7 e 3 5
8 d 2 10
9 a 3 5
N.B:- This question is related to Keeping the last N duplicates in pandas
df = df.groupby('Group').tail(3) is already an efficient way of doing it. Because you are overwriting the df variable, Python will take care of releasing the memory of the old dataframe, and you will only have access to the new one.
Trying way too hard to guess what you want.
NOTE: using Pandas inplace argument where it is available is NO guarantee that a new DataFrame won't be created in memory. In fact, it may very well create a new DataFrame in memory and replace the old one behind the scenes.
from collections import defaultdict
def f(s):
c = defaultdict(int)
for i, x in zip(s.index[::-1], s.values[::-1]):
c[x] += 1
if c[x] > 3:
yield i
df.drop([*f(df.Group)], inplace=True)
df
value Group something
0 a 1 1
1 b 1 2
2 c 1 4
5 x 2 5
6 d 2 3
7 e 3 5
8 d 2 10
9 a 3 5
Your answer already into the Post , However as earlier said in the comments you are overwriting the existing df , so to avoid that assign a new column name like below:
df['new_col'] = df.groupby('Group').tail(3)
However, out of curiosity, if you are not concerned about the the groupby and only looking for N last lines of the df yo can do it like below:
df[-2:] # last 2 rows
Related
When using groupby(), how can I create a DataFrame with a new column containing an index of the group number, similar to dplyr::group_indices in R. For example, if I have
>>> df=pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})
>>> df
a b
0 1 1
1 1 1
2 1 2
3 2 1
4 2 1
5 2 2
How can I get a DataFrame like
a b idx
0 1 1 1
1 1 1 1
2 1 2 2
3 2 1 3
4 2 1 3
5 2 2 4
(the order of the idx indexes doesn't matter)
Here is the solution using ngroup (available as of pandas 0.20.2) from a comment above by Constantino.
import pandas as pd
df = pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})
df['idx'] = df.groupby(['a', 'b']).ngroup()
df
a b idx
0 1 1 0
1 1 1 0
2 1 2 1
3 2 1 2
4 2 1 2
5 2 2 3
Here's a concise way using drop_duplicates and merge to get a unique identifier.
group_vars = ['a','b']
df.merge( df.drop_duplicates( group_vars ).reset_index(), on=group_vars )
a b index
0 1 1 0
1 1 1 0
2 1 2 2
3 2 1 3
4 2 1 3
5 2 2 5
The identifier in this case goes 0,2,3,5 (just a residual of original index) but this could be easily changed to 0,1,2,3 with an additional reset_index(drop=True).
Update: Newer versions of pandas (0.20.2) offer a simpler way to do this with the ngroup method as noted in a comment to the question above by #Constantino and a subsequent answer by #CalumYou. I'll leave this here as an alternate approach but ngroup seems like the better way to do this in most cases.
A simple way to do that would be to concatenate your grouping columns (so that each combination of their values represents a uniquely distinct element), then convert it to a pandas Categorical and keep only its labels:
df['idx'] = pd.Categorical(df['a'].astype(str) + '_' + df['b'].astype(str)).codes
df
a b idx
0 1 1 0
1 1 1 0
2 1 2 1
3 2 1 2
4 2 1 2
5 2 2 3
Edit: changed labels properties to codes as the former seem to be deprecated
Edit2: Added a separator as suggested by Authman Apatira
Definetely not the most straightforward solution, but here is what I would do (comments in the code):
df=pd.DataFrame({'a':[1,1,1,2,2,2],'b':[1,1,2,1,1,2]})
#create a dummy grouper id by just joining desired rows
df["idx"] = df[["a","b"]].astype(str).apply(lambda x: "".join(x),axis=1)
print df
That would generate an unique idx for each combination of a and b.
a b idx
0 1 1 11
1 1 1 11
2 1 2 12
3 2 1 21
4 2 1 21
5 2 2 22
But this is still a rather silly index (think about some more complex values in columns a and b. So let's clear the index:
# create a dictionary of dummy group_ids and their index-wise representation
dict_idx = dict(enumerate(set(df["idx"])))
# switch keys and values, so you can use dict in .replace method
dict_idx = {y:x for x,y in dict_idx.iteritems()}
#replace values with the generated dict
df["idx"].replace(dict_idx,inplace=True)
print df
That would produce the desired output:
a b idx
0 1 1 0
1 1 1 0
2 1 2 1
3 2 1 2
4 2 1 2
5 2 2 3
A way that I believe is faster than the current accepted answer by about an order of magnitude (timing results below):
def create_index_usingduplicated(df, grouping_cols=['a', 'b']):
df.sort_values(grouping_cols, inplace=True)
# You could do the following three lines in one, I just thought
# this would be clearer as an explanation of what's going on:
duplicated = df.duplicated(subset=grouping_cols, keep='first')
new_group = ~duplicated
return new_group.cumsum()
Timing results:
a = np.random.randint(0, 1000, size=int(1e5))
b = np.random.randint(0, 1000, size=int(1e5))
df = pd.DataFrame({'a': a, 'b': b})
In [6]: %timeit df['idx'] = pd.Categorical(df['a'].astype(str) + df['b'].astype(str)).codes
1 loop, best of 3: 375 ms per loop
In [7]: %timeit df['idx'] = create_index_usingduplicated(df, grouping_cols=['a', 'b'])
100 loops, best of 3: 17.7 ms per loop
I'm not sure this is such a trivial problem. Here is a somewhat convoluted solution that first sorts the grouping columns and then checks whether each row is different than the previous row and if so accumulates by 1. Check further below for an answer with string data.
df.sort_values(['a', 'b']).diff().fillna(0).ne(0).any(1).cumsum().add(1)
Output
0 1
1 1
2 2
3 3
4 3
5 4
dtype: int64
So breaking this up into steps, lets see the output of df.sort_values(['a', 'b']).diff().fillna(0) which checks if each row is different than the previous row. Any non-zero entry indicates a new group.
a b
0 0.0 0.0
1 0.0 0.0
2 0.0 1.0
3 1.0 -1.0
4 0.0 0.0
5 0.0 1.0
A new group only need to have a single column different so this is what .ne(0).any(1) checks - not equal to 0 for any of the columns. And then just a cumulative sum to keep track of the groups.
Answer for columns as strings
#create fake data and sort it
df=pd.DataFrame({'a':list('aabbaccdc'),'b':list('aabaacddd')})
df1 = df.sort_values(['a', 'b'])
output of df1
a b
0 a a
1 a a
4 a a
3 b a
2 b b
5 c c
6 c d
8 c d
7 d d
Take similar approach by checking if group has changed
df1.ne(df1.shift().bfill()).any(1).cumsum().add(1)
0 1
1 1
4 1
3 2
2 3
5 4
6 5
8 5
7 6
My current data frame is comprised of 10 rows and thousands of columns. The setup currently looks similar to this:
A B A B
1 2 3 4
5 6 7 8
But I desire something more like below, where essentially I would transpose the columns into rows once the headers start repeating themselves.
A B
1 2
5 6
3 4
7 8
I've been trying df.reshape but perhaps can't get the syntax right. Any suggestions on how best to transpose the data like this?
I'd probably go for stacking, grouping and then building a new DataFrame from scratch, eg:
pd.DataFrame({col: vals for col, vals in df.stack().groupby(level=1).agg(list).items()})
That'll also give you:
A B
0 1 2
1 3 4
2 5 6
3 7 8
Try with stack, groupby and pivot:
stacked = df.T.stack().to_frame().assign(idx=df.T.stack().groupby(level=0).cumcount()).reset_index()
output = stacked.pivot("idx", "level_0", 0).rename_axis(None, axis=1).rename_axis(None, axis=0)
>>> output
A B
0 1 2
1 5 6
2 3 4
3 7 8
I have a dataframe like this:
df1
a b c
0 1 2 [bg10, ng45, fg56]
1 4 5 [cv10, fg56]
2 7 8 [bg10, ng45, fg56]
3 7 8 [fg56, fg56]
4 7 8 [bg10]
I would like to count the total occurences take place of each type in column 'c'. I would then like to return the value of column 'b' for the values in column 'c' that have a count total of '1'.
The expected output is soemthing like this:
c b total_count
0 bg10 2 2
0 ng45 2 2
0 fg56 2 5
1 cv10 5 1
1 fg56 5 5
I have tried the 'Collections' library, and a 'for' loop (I understand its not best practise in Pandas) but i think i'm missing some fundamental udnerstanding of lists within cells, and how to perform analysis like these.
Thank you for taking my question into consideration.
I would use apply the following way:
first I create the df:
df1=pd.DataFrame({"b":[2,5,8,8], "c":[['bg10', 'ng45', 'fg56'],['cv10', 'fg56'],['bg10', 'ng45', 'fg56'],['fg56', 'fg56']]})
next use apply to count the number of (non unique) items in a list and save it in a different column:
df1["count_c"]=df1.c.apply(lambda x: len(x))
you will get the following:
b c count_c
0 2 [bg10, ng45, fg56] 3
1 5 [cv10, fg56] 2
2 8 [bg10, ng45, fg56] 3
3 8 [fg56, fg56] 2
to get the lines when c larger than threshold:`
df1[df1["count_c"]>2]["b"]
note: if you want to count only unique values in each list in column c you should use:
df1["count_c"]=df1.c.apply(lambda x: len(set(x)))
EDIT
in order to count the total number of each item I would try this:
first let's "unpack all the lists into columns
new_df1=(df1.c.apply(lambda x: pd.Series(x))).stack().reset_index(level=1,drop=True).to_frame("c").join(df1[["b"]],how="left")
then get the total counts of each item in the list and add it to a new col:
counts_dict=new_df1.c.value_counts().to_dict()
new_df1["total_count_c"]=new_df1.c.map(counts_dict)
new_df1.head()
c b total_count_c
0 bg10 2 2
0 ng45 2 2
0 fg56 2 5
1 cv10 5 1
1 fg56 5 5
I want to sort a subset of a dataframe (say, between indexes i and j) according to some value. I tried
df2=df.iloc[i:j].sort_values(by=...)
df.iloc[i:j]=df2
No problem with the first line but nothing happens when I run the second one (not even an error). How should I do ? (I tried also the update function but it didn't do either).
I believe need assign to filtered DataFrame with converting to numpy array by values for avoid align indices:
df = pd.DataFrame({'A': [1,2,3,4,3,2,1,4,1,2]})
print (df)
A
0 1
1 2
2 3
3 4
4 3
5 2
6 1
7 4
8 1
9 2
i = 2
j = 7
df.iloc[i:j] = df.iloc[i:j].sort_values(by='A').values
print (df)
A
0 1
1 2
2 1
3 2
4 3
5 3
6 4
7 4
8 1
9 2
Given this data frame:
import pandas as pd
df=pd.DataFrame({'A':[1,2,3],'B':[4,5,6],'C':[7,8,9]})
df
A B C
0 1 4 7
1 2 5 8
2 3 6 9
I'd like to create 3 new data frames; one from each column.
I can do this one at a time like this:
a=pd.DataFrame(df[['A']])
a
A
0 1
1 2
2 3
But instead of doing this for each column, I'd like to do it in a loop.
Here's what I've tried:
a=b=c=df.copy()
dfs=[a,b,c]
fields=['A','B','C']
for d,f in zip(dfs,fields):
d=pd.DataFrame(d[[f]])
...but when I then print each one, I get the whole original data frame as opposed to just the column of interest.
a
A B C
0 1 4 7
1 2 5 8
2 3 6 9
Update:
My actual data frame will have some columns that I do not need and the columns will not be in any sort of order, so I need to be able to get the columns by name.
Thanks in advance!
A simple list comprehension should be enough.
In [68]: df_list = [df[[x]] for x in df.columns]
Printing out the list, this is what you get:
In [69]: for d in df_list:
...: print(d)
...: print('-' * 5)
...:
A
0 1
1 2
2 3
-----
B
0 4
1 5
2 6
-----
C
0 7
1 8
2 9
-----
Each element in df_list is its own data frame, corresponding to each data frame from the original. Furthermore, you don't even need fields, use df.columns instead.
Or you can try this, instead create copy of df, this method will return the result as single Dataframe, not a list, However, I think save Dataframe into a list is better
dfs=['a','b','c']
fields=['A','B','C']
variables = locals()
for d,f in zip(dfs,fields):
variables["{0}".format(d)] = df[[f]]
a
Out[743]:
A
0 1
1 2
2 3
b
Out[744]:
B
0 4
1 5
2 6
c
Out[745]:
C
0 7
1 8
2 9
You should use loc
a = df.loc[:,0]
and then loop through like
for i in range(df.columns.size):
dfs[i] = df.loc[:, i]