So looking at this question a couple years later I see the error, to coerce the returned result so it assigns correctly you need to access the scalar values and use these to assign so they align as desired:

In [22]:
df.loc[df['A'] == 1, ['A', 'B']] = df['C'].values[0] + 10,df['C'].values[0] + 11
df

Out[22]:
    A   B  C
0  11  12  1
1   2   2  2
2   3   3  3

What I want to achieve: Condition: where column2 == 2 leave to be 2 if column1 < 30 elsif change to 3 if column1 > 90

This can be simplified into where (column2 == 2 and column1 > 90) set column2 to 3. The column1 < 30 part is redundant, since the value of column2 is only going to change from 2 to 3 if column1 > 90.

In the code that you provide, you are using pandas function replace, which operates on the entire Series, as stated in the reference:

Values of the Series are replaced with other values dynamically. This differs from updating with .loc or .iloc, which require you to specify a location to update with some value.

This means that for each iteration of for x in filter1 your code performs global replacement, which is not what you want to do - you want to update the specific row of column2 that corresponds to x from column1 (which you are iterating over).

the problem is 2 does not change to 3 where column1 > 90

This is truly strange. I would expect the code you provided to have changed every instance of 2 in column2 to 3 as soon as it encountered an x >= 30, as dictated by your code conditional statement (the execution of the else branch). This discrepancy may stem from the fact that you are assigning to column2 the result of global replacement performed on the column Output (the contents of which are unknown). In any case, if you want your program to do something under a specific condition, such as x > 90, it should be explicitly stated in the code. You should also note that the statement data['column2'] = data['column2'].replace([2], [2]) achieves nothing, since 2 is being replaced with 2 and the same column is both the source and the destination.

What you could use to solve this particular task is a boolean mask (or the query method). Both are explained in an excellent manner in this question.

Using a boolean mask would be the easiest approach in your case:

mask = (data['column2'] == 2) & (data['column1'] > 90)
data['column2'][mask] = 3

The first line builds a Series of booleans (True/False) that indicate whether the supplied condition is satisfied. The second line assigns the value 3 to those rows of column2 where the mask is True.

Using:

df[ (df.A=='blue') & (df.B=='red') & (df.C=='square') ]['D'] = 'succeed'

gives the warning:

/usr/local/lib/python2.7/dist-packages/ipykernel_launcher.py:2: 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

A better way of achieving this seems to be:

df.loc[(df['A'] == 'blue') & (df['B'] == 'red') & (df['C'] == 'square'),'D'] = 'M5'

you want to replace

print df.loc[df['Col1'].isnull(),['Col1','Col2', 'Col3']]

  Col1 Col2 Col3
2  NaN  NaN  NaN
3  NaN  NaN  NaN

With:

replace_with_this = df.loc[df['Col1'].isnull(),['col1_v2','col2_v2', 'col3_v2']]
print replace_with_this

  col1_v2 col2_v2 col3_v2
2       a       b       d
3       d       e       f

Seems reasonable. However, when you do the assignment, you need to account for index alignment, which includes columns.

So, this should work:

df.loc[df['Col1'].isnull(),['Col1','Col2', 'Col3']] = replace_with_this.values

print df

  Col1 Col2 Col3 col1_v2 col2_v2 col3_v2
0    A    B    C     NaN     NaN     NaN
1    D    E    F     NaN     NaN     NaN
2    a    b    d       a       b       d
3    d    e    f       d       e       f

I accounted for columns by using .values at the end. This stripped the column information from the replace_with_this dataframe and just used the values in the appropriate positions.

I think you can use loc if you need update two columns to same value:

df1.loc[df1['stream'] == 2, ['feat','another_feat']] = 'aaaa'
print df1
   stream        feat another_feat
a       1  some_value   some_value
b       2        aaaa         aaaa
c       2        aaaa         aaaa
d       3  some_value   some_value

If you need update separate, one option is use:

df1.loc[df1['stream'] == 2, 'feat'] = 10
print df1
   stream        feat another_feat
a       1  some_value   some_value
b       2          10   some_value
c       2          10   some_value
d       3  some_value   some_value

Another common option is use numpy.where:

df1['feat'] = np.where(df1['stream'] == 2, 10,20)
print df1
   stream  feat another_feat
a       1    20   some_value
b       2    10   some_value
c       2    10   some_value
d       3    20   some_value

EDIT: If you need divide all columns without stream where condition is True, use:

print df1
   stream  feat  another_feat
a       1     4             5
b       2     4             5
c       2     2             9
d       3     1             7

#filter columns all without stream
cols = [col for col in df1.columns if col != 'stream']
print cols
['feat', 'another_feat']

df1.loc[df1['stream'] == 2, cols ] = df1 / 2
print df1
   stream  feat  another_feat
a       1   4.0           5.0
b       2   2.0           2.5
c       2   1.0           4.5
d       3   1.0           7.0

If working with multiple conditions is possible use multiple numpy.where or numpy.select:

df0 = pd.DataFrame({'Col':[5,0,-6]})

df0['New Col1'] = np.where((df0['Col'] > 0), 'Increasing', 
                          np.where((df0['Col'] < 0), 'Decreasing', 'No Change'))

df0['New Col2'] = np.select([df0['Col'] > 0, df0['Col'] < 0],
                            ['Increasing',  'Decreasing'], 
                            default='No Change')

print (df0)
   Col    New Col1    New Col2
0    5  Increasing  Increasing
1    0   No Change   No Change
2   -6  Decreasing  Decreasing

You can use np.select for this:

import pandas as pd
import numpy as np

df = pd.DataFrame({ 'a': ['one', 'one', 'three', 'two', 'eleven', 'two'],
              'b': [45, 34, 556, 32, 97, 33],
              'c': [234, 66, 12, 44, 99, 3],
              'd': [123, 45, 55, 98, 17, 22] })

df['e'] = df.b + df.c + df.d

# list with your conditions
conditions = [(df.a == 'one') & (df.b < 50),
              (df.a == 'two') & (df.d > 50)]

# list with accompanying choices
choices = [0,1]

df['f'] = np.select(conditions, choices, 2) 
# 2 being the default: i.e. the 'else' choice.

df

        a    b    c    d    e  f
0     one   45  234  123  402  0
1     one   34   66   45  145  0
2   three  556   12   55  623  2
3     two   32   44   98  174  1
4  eleven   97   99   17  213  2
5     two   33    3   22   58  2

You access string functions with .str. This should work:

df.loc[df.Lvalue.str.istitle(), 'IsLCap'] = 1

Update To check the length of the strings in the column you can use the string method .str.len:

df.loc[df.Lvalue.str.len() < 3, 'less_than_three'] = 1

Pandas Series string methods

This will work

df.loc[(df.Color=='Blue')&(df.Age==28)&(df.City=='Atl'),'Value']=1
df
Out[687]: 
   Color Name  Age City  Value
0   Blue  Bob   28  Atl      1
1  Green  Bob   27  Chi      0
2   Blue  Sam   28  Atl      1

I did it like this: where : A-B = -B+A

df[df> 100000] = -1*df[df> 100000] + 4294967295+1