I have the following dataframe:
df = pd.DataFrame([[np.nan, 2, np.nan, 0],
[3, 4, np.nan, 1],
[np.nan, np.nan, 5, np.nan],
[np.nan, 3, np.nan, 4]],
columns=list('ABCD'))
I want to do a ffill() on column B with df["B"].ffill(inplace=True) which results in the following df:
A B C D
0 NaN 2.0 NaN 0.0
1 3.0 4.0 NaN 1.0
2 NaN 4.0 5.0 NaN
3 NaN 3.0 NaN 4.0
Now I want to replace all NaN values with their corresponding value from column B. The documentation states that you can give fillna() a Series, so I tried df.fillna(df["B"], inplace=True). This results in the exact same dataframe as above.
However, if I put in a simple value (e.g. df.fillna(0, inplace=True), then it does work:
A B C D
0 0.0 2.0 0.0 0.0
1 3.0 4.0 0.0 1.0
2 0.0 4.0 5.0 0.0
3 0.0 3.0 0.0 4.0
The funny thing is that the fillna() does seem to work with a Series as value parameter when operated on another Series object. For example, df["A"].fillna(df["B"], inplace=True) results in:
A B C D
0 2.0 2.0 NaN 0
1 3.0 4.0 NaN 1
2 4.0 4.0 NaN 5
3 3.0 3.0 NaN 4
My real dataframe has a lot of columns and I would hate to manually fillna() all of them. Am I overlooking something here? Didn't I understand the docs correctly perhaps?
EDIT I have clarified my example in such a way that 'ffill' with axis=1 does not work for me. In reality, my dataframe has many, many columns (hundreds) and I am looking for a way to not have to explicitly mention all the columns.
Try changing the axis to 1 (columns):
df = df.ffill(1).bfill(1)
If you need to specify the columns, you can do something like this:
df[["B","C"]] = df[["B","C"]].ffill(1)
EDIT:
Since you need something more general and df.fillna(df.B, axis = 1) is not implemented yet, you can try with:
df = df.T.fillna(df.B).T
Or, equivalently:
df.T.fillna(df.B, inplace=True)
This works because the indices of df.B coincides with the columns of df.T so pandas will know how to replace it. From the docs:
value: scalar, dict, Series, or DataFrame.
Value to use to fill holes (e.g. 0), alternately a dict/Series/DataFrame of values specifying which value to use for each index (for a Series) or column (for a DataFrame). Values not in the dict/Series/DataFrame will not be filled. This value cannot be a list.
So, for example, the NaN in column 0 at row A (in df.T) will be replaced for the value with index 0 in df.B.
Related
Using df.compare in Pandas, is it possible to change the labels of self/other from the output?
I need to send this output directly to less technically savvy users and would like to change them to more descriptive labels.
My code:
if df_1.equals(df_2):
return None
else:
return df_1.compare(df_2, align_axis=0)
You can rename the index level to something more obvious:
df1 = pd.DataFrame([[1,2,3,4], [1,2,3,4]])
df2 = pd.DataFrame([[1,2,5,4], [5,2,3,1]])
df1.compare(df2, align_axis=0).rename(index={'self': 'left', 'other': 'right'}, level=-1)
0 2 3
0 left NaN 3.0 NaN
right NaN 5.0 NaN
1 left 1.0 NaN 4.0
right 5.0 NaN 1.0
How do I combine values from two rows with identical index and has no intersection in values?
import pandas as pd
df = pd.DataFrame([[1,2,3],[4,None,None],[None,5,6]],index=['a','b','b'])
df
#input
0 1 2
a 1.0 2.0 3.0
b 4.0 NaN NaN
b NaN 5.0 6.0
Desired output
0 1 2
a 1.0 2.0 3.0
b 4.0 5.0 6.0
Please stack(), drops all nans and unstack()
df.stack().unstack()
If possible simplify solution for first non missing values per index labels use GroupBy.first:
df1 = df.groupby(level=0).first()
If possible same output from sample data is use sum per labels use sum:
df1 = df.sum(level=0)
If there is multiple non missing values per groups is necessary specify expected output, obviously is is more complicated.
Is there a way to replace NAN values in both categorical columns as well as numerical columns at once?
A very simplistic example:
data = {'col_1': [3, np.nan, 1, 2], 'col_2': ['a', 'a', np.nan, 'd']}
df = pd.DataFrame.from_dict(data)
Dataframe:
col_1 col_2
0 3.0 a
1 NaN a
2 1.0 NaN
3 0.0 d
Goal:
To replace col_1's NAN with the mean of col_1 and replace col_2's NAN with the mode ('a') of col_2.
Right now, I have to replace it for each column individually. If all columns are numeric or categorical then it's easy because the operation can be applied on the whole data frame but I couldn't find a way to do it one line for a mixed data frame.
mean will only work for numeric types, so fill that first then fill the remainder with mode.
df.fillna(df.mean()).fillna(df.mode().iloc[0])
# col_1 col_2
#0 3.0 a
#1 2.0 a
#2 1.0 a
#3 2.0 d
If you have ties, the mode will be the one that is sorted first.
What I will do
df.fillna(df.agg(['mean',lambda x : x.value_counts().index[0]]).ffill().iloc[-1,:])
col_1 col_2
0 3.0 a
1 2.0 a
2 1.0 a
3 2.0 d
I have a dataframe column which contains a list of numbers from a .csv. These numbers range from 1-1400 and may or may not be repeated and the a NaN value can appear pretty much anywhere at random.
Two examples would be
a=[1,4,NaN,5,6,7,...1398,1400,1,2,3,NaN,8,9,...,1398,NaN]
b=[1,NaN,2,3,4,NaN,7,10,...,1398,1399,1400]
I would like to create another column that finds the first 1-1400 and records a '1' in the same index and if the second set of 1-1400 exists, then mark that down as a '2' in the new column
I can think of some roundabout ways using temporary placeholders and some other kind of checks, but I was wondering if there was a 1-3 liner to do this operation
Edit1: I would prefer there to be a single column returned
a1=[1,1,NaN,1,1,1,...1,1,2,2,2,NaN,2,2,...,2,NaN]
b1=[1,NaN,1,1,1,NaN,1,1,...,1,1,1]
You can use groupby() and cumcount() to count numbers in each column:
# create new columns for counting
df['a1'] = np.nan
df['b1'] = np.nan
# take groupby for each value in column `a` and `b` and count each value
df.a1 = df.groupby('a').cumcount() + 1
df.b1 = df.groupby('b').cumcount() + 1
# set np.nan as it is
df.loc[df.a.isnull(), 'a1'] = np.nan
df.loc[df.b.isnull(), 'b1'] = np.nan
EDIT (after receiving a comment of 'does not work'):
df['a2'] = df.ffill().a.diff()
df['a1'] = df.loc[df.a2 < 0].groupby('a').cumcount() + 1
df['a1'] = df['a1'].bfill().shift(-1)
df.loc[df.a1.isnull(), 'a1'] = df.a1.max() + 1
df.drop('a2', axis=1, inplace=True)
df.loc[df.a.isnull(), 'a1'] = np.nan
you can use diff to check when the difference between two following values is negative, meaning of the start of a new range. Let's create a dataframe:
import pandas as pd
import numpy as np
# to create a dataframe with two columns my range go up to 12 but 1400 is the same
df = pd.DataFrame({'a':[1,4,np.nan,5,10,12,2,3,4,np.nan,8,12],'b':range(1,13)})
df.loc[[4,8],'b'] = np.nan
Because you have 'NaN', you need to use ffill to fill NaN with previous value and you want the opposite of the row (using ~) where the diff is greater or equal than 0 (I know it sound like less than 0, but not exactely here as it miss the first row of the dataframe). For column 'a' for example
print (df.loc[~(df.a.ffill().diff()>=0),'a'])
0 1.0
6 2.0
Name: a, dtype: float64
you get the two rows where a "new" range start. To use this property to create 'a1', you can do:
# put 1 in the rows with a new range start
df.loc[~(df.a.ffill().diff()>=0),'a1'] = 1
# create a mask to select notnull row in a:
mask_a = df.a.notnull()
# use cumsum and ffill on column a1 with the mask_a
df.loc[mask_a,'a1'] = df.loc[mask_a,'a1'].cumsum().ffill()
Finally, for several column, you can do:
list_col = ['a','b']
for col in list_col:
df.loc[~(df[col].ffill().diff()>=0),col+'1'] = 1
mask = df[col].notnull()
df.loc[mask,col+'1'] = df.loc[mask,col+'1'].cumsum().ffill()
and with my input, you get:
a b a1 b1
0 1.0 1.0 1.0 1.0
1 4.0 2.0 1.0 1.0
2 NaN 3.0 NaN 1.0
3 5.0 4.0 1.0 1.0
4 10.0 NaN 1.0 NaN
5 12.0 6.0 1.0 1.0
6 1.0 7.0 2.0 1.0
7 3.0 8.0 2.0 1.0
8 4.0 NaN 2.0 NaN
9 NaN 10.0 NaN 1.0
10 8.0 11.0 2.0 1.0
11 12.0 12.0 2.0 1.0
EDIT: you can even do it in one line for each column, same result:
df['a1'] = df[df.a.notnull()].a.diff().fillna(-1).lt(0).cumsum()
df['b1'] = df[df.b.notnull()].b.diff().fillna(-1).lt(0).cumsum()
my code:
import pandas as pd
from sklearn.preprocessing import LabelEncoder
column_names = ["age","workclass","fnlwgt","education","education-num","marital-status","occupation","relationship","race","sex","capital-gain","capital-loss","hrs-per-week","native-country","income"]
adult_train = pd.read_csv("adult.data",header=None,sep=',\s',na_values=["?"])
adult_train.columns=column_names
adult_train.fillna('NA',inplace=True)
I want the index of the rows which have the value 'NA' in more than one column. Is there an inbuilt method or I have to iterate row wise and check values at each column?
here is the snapshot of the data:
I want index of rows like 398,409(missing values at column B and G) and not of rows like 394(missing value only at column N)
Use isnull.any(1) or sum to get the boolean mask, then select the rows to get the index i.e
df = pd.DataFrame({'A':[1,2,3,4,5],
'B' :[np.nan,4,5,np.nan,8],
'C' :[2,4,np.nan,3,5],
'D' :[np.nan,np.nan,np.nan,np.nan,5]})
A B C D
0 1 NaN 2.0 NaN
1 2 4.0 4.0 NaN
2 3 5.0 NaN NaN
3 4 NaN 3.0 NaN
4 5 8.0 5.0 5.0
# If you want to select rows with nan value from Columns B and C
df.loc[df[['B','C']].isnull().any(1)].index
Int64Index([0, 2, 3], dtype='int64')
# If you want to rows with more than one nan then
df.loc[df.isnull().sum(1)>1].index
Int64Index([0, 2, 3], dtype='int64')