I have a big dataframe with many duplicates in it. I want to keep the first and last entry of each duplicate but drop every duplicate in between.
I've already tried to get this done by using df.drop_duplicates with the parameters 'first' and 'last' to get two dataframes and then merge them again to one df so I have the first and last entry, but that didn't work.
df_first = df
df_last = df
df_first['Path'].drop_duplicates(keep='first', inplace=True)
df_last['Path'].drop_duplicates(keep='last', inplace=True)
Thanks for your help in advance!
Use GroupBy.nth for avoid duplicates if group with length is 1:
df = pd.DataFrame({
'a':[5,3,6,9,2,4],
'Path':list('aaabbc')
})
print(df)
a Path
0 5 a
1 3 a
2 6 a
3 9 b
4 2 b
5 4 c
df = df.groupby('Path').nth([0, -1])
print (df)
a
Path
a 5
a 6
b 9
b 2
c 4
**Using group by.nth which is an Updated code from previous solution to get nth entry
def keep_second_dup(duplicate):
duplicate[Columnname]=duplicate[Columnname'].value_counts()
second_duplicate=duplicate[duplicate['Count']>=1]
residual=duplicate[duplicate['Count']==1]
sec=second_duplicated.groupby([Columnname]).nth([1]).reset_index()
final_data=pd.concat([sec,residual])
final_data.drop('Count',axis=1,inplace=True)
return final_data
Related
I have some troubles with my Python work,
my steps are:
1)add the list to ordinary Dataframe
2)delete the columns which is min in the list
my list is called 'each_c' and my ordinary Dataframe is called 'df_col'
I want it to become like this:
hope someone can help me, thanks!
This is clearly described in the documentation: https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.drop.html
df_col.drop(columns=[3])
Convert each_c to Series, append by DataFrame.append and then get indices by minimal value by Series.idxmin and pass to drop - it remove only first minimal column:
s = pd.Series(each_c)
df = df_col.append(s, ignore_index=True).drop(s.idxmin(), axis=1)
If need remove all columns if multiple minimals:
each_c = [-0.025,0.008,-0.308,-0.308]
s = pd.Series(each_c)
df_col = pd.DataFrame(np.random.random((10,4)))
df = df_col.append(s, ignore_index=True)
df = df.loc[:, s.ne(s.min())]
print (df)
0 1
0 0.602312 0.641220
1 0.586233 0.634599
2 0.294047 0.339367
3 0.246470 0.546825
4 0.093003 0.375238
5 0.765421 0.605539
6 0.962440 0.990816
7 0.810420 0.943681
8 0.307483 0.170656
9 0.851870 0.460508
10 -0.025000 0.008000
EDIT: If solution raise error:
IndexError: Boolean index has wrong length:
it means there is no default columns name by range - 0,1,2,3. Possible solution is set index values in Series by rename:
each_c = [-0.025,0.008,-0.308,-0.308]
df_col = pd.DataFrame(np.random.random((10,4)), columns=list('abcd'))
s = pd.Series(each_c).rename(dict(enumerate(df.columns)))
df = df_col.append(s, ignore_index=True)
df = df.loc[:, s.ne(s.min())]
print (df)
a b
0 0.321498 0.327755
1 0.514713 0.575802
2 0.866681 0.301447
3 0.068989 0.140084
4 0.069780 0.979451
5 0.629282 0.606209
6 0.032888 0.204491
7 0.248555 0.338516
8 0.270608 0.731319
9 0.732802 0.911920
10 -0.025000 0.008000
I am trying to select the first 2 columns and the last 2 column from a data frame by index with pandas and save it on the same dataframe.
is there a way to do that in one step?
You can use the iloc function to get the columns, and then pass in the indexes.
df.iloc[:,[0,1,-1,-2]]
You are looking for iloc:
df = pd.DataFrame([[1,2,3,4,5], [2,3,4,5,6], [3,4,5,6,7]], columns=['a','b','c','d','e'])
df.iloc[:,:2] # Grabs all rows and first 2 columns
df.iloc[:,-2:] # Grabs all rows and last 2 columns
pd.concat([df.iloc[:,:2],df.iloc[:,-2:]],axis=1) # Puts them together row wise
df = pd.DataFrame([[1,2,3,4,5], [2,3,4,5,6], [3,4,5,6,7]], columns=['a','b','c','d','e'])
df[['a','b','d','e']]
result
a b d e
0 1 2 4 5
1 2 3 5 6
2 3 4 6 7
I have two pandas dataframes with names df1 and df2 such that
`
df1: a b c d
1 2 3 4
5 6 7 8
and
df2: b c
12 13
I want the result be like
result: b c
2 3
6 7
Here it should be noted that a b c d are the column names in pandas dataframe. The shape and values of both pandas dataframe are different. I want to match the column names of df2 with that of column names of df1 and select all the rows of df1 the headers of which are matched with the column names of df2.. df2 is only used to select the specific columns of df1 maintaining all the rows. I tried some code given below but that gives me an empty index.
df1.columns.intersection(df2.columns)
The above code is not giving me my resut as it gives index headers with no values. I want to write a code in which I can give my two dataframes as input and it compares the columns headers for selection. I don't have to hard code column names.
I believe you need:
df = df1[df1.columns.intersection(df2.columns)]
Or like #Zero pointed in comments:
df = df1[df1.columns & df2.columns]
Or, use reindex
In [594]: df1.reindex(columns=df2.columns)
Out[594]:
b c
0 2 3
1 6 7
Also as
In [595]: df1.reindex(df2.columns, axis=1)
Out[595]:
b c
0 2 3
1 6 7
Alternatively to intersection:
df = df1[df1.columns.isin(df2.columns)]
Let's say I have a DataFrame that looks like this:
a b c d e f g
1 2 3 4 5 6 7
4 3 7 1 6 9 4
8 9 0 2 4 2 1
How would I go about deleting every column besides a and b?
This would result in:
a b
1 2
4 3
8 9
I would like a way to delete these using a simple line of code that says, delete all columns besides a and b, because let's say hypothetically I have 1000 columns of data.
Thank you.
In [48]: df.drop(df.columns.difference(['a','b']), 1, inplace=True)
Out[48]:
a b
0 1 2
1 4 3
2 8 9
or:
In [55]: df = df.loc[:, df.columns.intersection(['a','b'])]
In [56]: df
Out[56]:
a b
0 1 2
1 4 3
2 8 9
PS please be aware that the most idiomatic Pandas way to do that was already proposed by #Wen:
df = df[['a','b']]
or
df = df.loc[:, ['a','b']]
Another option to add to the mix. I prefer this approach for readability.
df = df.filter(['a', 'b'])
Where the first positional argument is items=[]
Bonus
You can also use a like argument or regex to filter.
Helpful if you have a set of columns like ['a_1','a_2','b_1','b_2']
You can do
df = df.filter(like='b_')
and end up with ['b_1','b_2']
Pandas documentation for filter.
there are multiple solution .
df = df[['a','b']] #1
df = df[list('ab')] #2
df = df.loc[:,df.columns.isin(['a','b'])] #3
df = pd.DataFrame(data=df.eval('a,b').T,columns=['a','b']) #4 PS:I do not recommend this method , but still a way to achieve this
Hey what you are looking for is:
df = df[["a","b"]]
You will recive a dataframe which only contains the columns a and b
If you only want to keep more columns than you're dropping put a "~" before the .isin statement to select every column except the ones you want:
df = df.loc[:, ~df.columns.isin(['a','b'])]
If you have more than two columns that you want to drop, let's say 20 or 30, you can use lists as well. Make sure that you also specify the axis value.
drop_list = ["a","b"]
df = df.drop(df.columns.difference(drop_list), axis=1)
I have a the following data frame:
I want to remove duplicate data in WD column, if they have the same drug_id.
For example, there is two "crying" in WD column with the same drug_id = 32. So I want to remove one of the row that has crying.
How I can do it? I know how to duplicate rows, but I do not know how to add this condition to this code.
df = df.apply(lambda x:x.drop_duplicates())
You can use drop_duplicates with subset parameter which optionally considers certain columns for duplicates:
df.drop_duplicates(subset = ["drug_id", "WD"])
If the upper/lower cases are important for considering duplicates, you could try:
df[~df[['drug_id', 'WD']].apply(lambda x: x.str.lower()).duplicated()]
Where you can convert both drug_id and WD columns to lower case, use duplicated() method to identify duplicated rows and then use the generated logical series to filter out duplicated rows.
Example:
df = pd.DataFrame({"A": [1,1,2,2], "B":[1,2,3,4], "C":[1,1,2,3]})
df
# A B C
#0 1 1 1
#1 1 2 1
#2 2 3 2
#3 2 4 3
df.drop_duplicates(subset=['A', 'C'])
# A B C
#0 1 1 1
#2 2 3 2
#3 2 4 3