Hi I want to remove all duplicates rows from panda dataframe by only keeping first. This is what i am doing.
import pandas as pd
df = pd.DataFrame({'col1':['A']*3+['B']*4+['C','B','A'],'col2':[2,3,4,2,4,2,1,3,4,4]})
print(df)
df.drop_duplicates(subset=None, keep='first', inplace=True, ignore_index=True)
This is fine but the given solution is exceeding time limit in my system. Can someone provide a better solution?
I expect it to be much quicker with NumPy:
>>> pd.DataFrame(np.unique(df.to_numpy(dtype=str), axis=0), columns=df.columns)
col1 col2
0 A 2
1 A 3
2 A 4
3 B 1
4 B 2
5 B 4
6 C 3
>>>
Using np.unique.
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 am merging two data frames using pandas.merge. Even after specifying how = left option, I found the number of rows of merged data frame is larger than the original. Why does this happen?
panel = pd.read_csv(file1, encoding ='cp932')
before_len = len(panel)
prof_2000 = pd.read_csv(file2, encoding ='cp932').drop_duplicates()
temp_2000 = pd.merge(panel, prof_2000, left_on='Candidate_u', right_on="name2", how="left")
after_len = len(temp_2000)
print(before_len, after_len)
> 12661 13915
This sounds like having more than one rows in right under 'name2' that match the key you have set for the left. Using option 'how='left' with pandas.DataFrame.merge() only means that:
left: use only keys from left frame
However, the actual number of rows in the result object is not necessarily going to be the same as the number of rows in the left object.
Example:
In [359]: df_1
Out[359]:
A B
0 a AAA
1 b BBA
2 c CCF
and then another DF that looks like this (notice that there are more than one entry for your desired key on the left):
In [360]: df_3
Out[360]:
key value
0 a 1
1 a 2
2 b 3
3 a 4
If I merge these two on left.A, here's what happens:
In [361]: df_1.merge(df_3, how='left', left_on='A', right_on='key')
Out[361]:
A B key value
0 a AAA a 1.0
1 a AAA a 2.0
2 a AAA a 4.0
3 b BBA b 3.0
4 c CCF NaN NaN
This happened even though I merged with how='left' as you can see above, there were simply more than one rows to merge and as shown here the result pd.DataFrame has in fact more rows than the pd.DataFrame on the left.
I hope this helps!
The problem of doubling of rows after each merge() (of any type, 'both' or 'left') is usually caused by duplicates in any of the keys, so we need to drop them first:
left_df.drop_duplicates(subset=left_key, inplace=True)
right_df.drop_duplicates(subset=right_key, inplace=True)
If you do not have any duplication, as indicated in the above answer. You should double-check the names of removed entries. In my case, I discovered that the names of removed entries are inconsistent between the df1 and df2 and I solved the problem by:
df1["col1"] = df2["col2"]
So I created two dataframes from existing CSV files, both consisting of entirely numbers. The second dataframe consists of an index from 0 to 8783 and one column of numbers and I want to add it on as a new column to the first dataframe which has an index consisting of a month, day and hour. I tried using append, merge and concat and none worked and then tried simply using:
x1GBaverage['Power'] = x2_cut
where x1GBaverage is the first dataframe and x2_cut is the second. When I did this it added x2_cut on properly but all the values were entered as NaN instead of the numerical values that they should be. How should I be approaching this?
x1GBaverage['Power'] = x2_cut.values
problem solved :)
The thing about pandas is that values are implicitly linked to their indices unless you deliberately specify that you only need the values to be transferred over.
If they're the same row counts and you just want to tack it on the end, the indexes either need to match, or you need to just pass the underlying values. In the example below, columns 3 and 5 are the index matching & value versions, and 4 is what you're running into now:
In [58]: df = pd.DataFrame(np.random.random((3,3)))
In [59]: df
Out[59]:
0 1 2
0 0.670812 0.500688 0.136661
1 0.185841 0.239175 0.542369
2 0.351280 0.451193 0.436108
In [61]: df2 = pd.DataFrame(np.random.random((3,1)))
In [62]: df2
Out[62]:
0
0 0.638216
1 0.477159
2 0.205981
In [64]: df[3] = df2
In [66]: df.index = ['a', 'b', 'c']
In [68]: df[4] = df2
In [70]: df[5] = df2.values
In [71]: df
Out[71]:
0 1 2 3 4 5
a 0.670812 0.500688 0.136661 0.638216 NaN 0.638216
b 0.185841 0.239175 0.542369 0.477159 NaN 0.477159
c 0.351280 0.451193 0.436108 0.205981 NaN 0.205981
If the row counts differ, you'll need to use df.merge and let it know which columns it should be using to join the two frames.
What's the easiest way to add an empty column to a pandas DataFrame object? The best I've stumbled upon is something like
df['foo'] = df.apply(lambda _: '', axis=1)
Is there a less perverse method?
If I understand correctly, assignment should fill:
>>> import numpy as np
>>> import pandas as pd
>>> df = pd.DataFrame({"A": [1,2,3], "B": [2,3,4]})
>>> df
A B
0 1 2
1 2 3
2 3 4
>>> df["C"] = ""
>>> df["D"] = np.nan
>>> df
A B C D
0 1 2 NaN
1 2 3 NaN
2 3 4 NaN
To add to DSM's answer and building on this associated question, I'd split the approach into two cases:
Adding a single column: Just assign empty values to the new columns, e.g. df['C'] = np.nan
Adding multiple columns: I'd suggest using the .reindex(columns=[...]) method of pandas to add the new columns to the dataframe's column index. This also works for adding multiple new rows with .reindex(rows=[...]). Note that newer versions of Pandas (v>0.20) allow you to specify an axis keyword rather than explicitly assigning to columns or rows.
Here is an example adding multiple columns:
mydf = mydf.reindex(columns = mydf.columns.tolist() + ['newcol1','newcol2'])
or
mydf = mydf.reindex(mydf.columns.tolist() + ['newcol1','newcol2'], axis=1) # version > 0.20.0
You can also always concatenate a new (empty) dataframe to the existing dataframe, but that doesn't feel as pythonic to me :)
I like:
df['new'] = pd.Series(dtype='int')
# or use other dtypes like 'float', 'object', ...
If you have an empty dataframe, this solution makes sure that no new row containing only NaN is added.
Specifying dtype is not strictly necessary, however newer Pandas versions produce a DeprecationWarning if not specified.
an even simpler solution is:
df = df.reindex(columns = header_list)
where "header_list" is a list of the headers you want to appear.
any header included in the list that is not found already in the dataframe will be added with blank cells below.
so if
header_list = ['a','b','c', 'd']
then c and d will be added as columns with blank cells
Starting with v0.16.0, DF.assign() could be used to assign new columns (single/multiple) to a DF. These columns get inserted in alphabetical order at the end of the DF.
This becomes advantageous compared to simple assignment in cases wherein you want to perform a series of chained operations directly on the returned dataframe.
Consider the same DF sample demonstrated by #DSM:
df = pd.DataFrame({"A": [1,2,3], "B": [2,3,4]})
df
Out[18]:
A B
0 1 2
1 2 3
2 3 4
df.assign(C="",D=np.nan)
Out[21]:
A B C D
0 1 2 NaN
1 2 3 NaN
2 3 4 NaN
Note that this returns a copy with all the previous columns along with the newly created ones. In order for the original DF to be modified accordingly, use it like : df = df.assign(...) as it does not support inplace operation currently.
if you want to add column name from a list
df=pd.DataFrame()
a=['col1','col2','col3','col4']
for i in a:
df[i]=np.nan
df["C"] = ""
df["D"] = np.nan
Assignment will give you this warning SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame. Try
using .loc[row_indexer,col_indexer] = value instead
so its better to use insert:
df.insert(index, column-name, column-value)
#emunsing's answer is really cool for adding multiple columns, but I couldn't get it to work for me in python 2.7. Instead, I found this works:
mydf = mydf.reindex(columns = np.append( mydf.columns.values, ['newcol1','newcol2'])
One can use df.insert(index_to_insert_at, column_header, init_value) to insert new column at a specific index.
cost_tbl.insert(1, "col_name", "")
The above statement would insert an empty Column after the first column.
this will also work for multiple columns:
df = pd.DataFrame({"A": [1,2,3], "B": [2,3,4]})
>>> df
A B
0 1 2
1 2 3
2 3 4
df1 = pd.DataFrame(columns=['C','D','E'])
df = df.join(df1, how="outer")
>>>df
A B C D E
0 1 2 NaN NaN NaN
1 2 3 NaN NaN NaN
2 3 4 NaN NaN NaN
Then do whatever you want to do with the columns
pd.Series.fillna(),pd.Series.map()
etc.
The below code address the question "How do I add n number of empty columns to my existing dataframe". In the interest of keeping solutions to similar problems in one place, I am adding it here.
Approach 1 (to create 64 additional columns with column names from 1-64)
m = list(range(1,65,1))
dd=pd.DataFrame(columns=m)
df.join(dd).replace(np.nan,'') #df is the dataframe that already exists
Approach 2 (to create 64 additional columns with column names from 1-64)
df.reindex(df.columns.tolist() + list(range(1,65,1)), axis=1).replace(np.nan,'')
You can do
df['column'] = None #This works. This will create a new column with None type
df.column = None #This will work only when the column is already present in the dataframe
If you have a list of columns that you want to be empty, you can use assign, then comprehension dict, then dict unpacking.
>>> df = pd.DataFrame({"A": [1,2,3], "B": [2,3,4]})
>>> nan_cols_name = ["C","D","whatever"]
>>> df.assign(**{col:np.nan for col in nan_cols_name})
A B C D whatever
0 1 2 NaN NaN NaN
1 2 3 NaN NaN NaN
2 3 4 NaN NaN NaN
You can also unpack multiple dict in a dict that you unpack if you want different values for different columns.
df = pd.DataFrame({"A": [1,2,3], "B": [2,3,4]})
nan_cols_name = ["C","D","whatever"]
empty_string_cols_name = ["E","F","bad column with space"]
df.assign(**{
**{col:np.nan for col in my_empy_columns_name},
**{col:"" for col in empty_string_cols_name}
}
)
Sorry for I did not explain my answer really well at beginning. There is another way to add an new column to an existing dataframe.
1st step, make a new empty data frame (with all the columns in your data frame, plus a new or few columns you want to add) called df_temp
2nd step, combine the df_temp and your data frame.
df_temp = pd.DataFrame(columns=(df_null.columns.tolist() + ['empty']))
df = pd.concat([df_temp, df])
It might be the best solution, but it is another way to think about this question.
the reason of I am using this method is because I am get this warning all the time:
: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
df["empty1"], df["empty2"] = [np.nan, ""]
great I found the way to disable the Warning
pd.options.mode.chained_assignment = None
The reason I was looking for such a solution is simply to add spaces between multiple DFs which have been joined column-wise using the pd.concat function and then written to excel using xlsxwriter.
df[' ']=df.apply(lambda _: '', axis=1)
df_2 = pd.concat([df,df1],axis=1) #worked but only once.
# Note: df & df1 have the same rows which is my index.
#
df_2[' ']=df_2.apply(lambda _: '', axis=1) #didn't work this time !!?
df_4 = pd.concat([df_2,df_3],axis=1)
I then replaced the second lambda call with
df_2['']='' #which appears to add a blank column
df_4 = pd.concat([df_2,df_3],axis=1)
The output I tested it on was using xlsxwriter to excel.
Jupyter blank columns look the same as in excel although doesnt have xlsx formatting.
Not sure why the second Lambda call didnt work.