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Pandas/Python Looping through rows of a data frame and change one column based on another column value that contains NA'a [duplicate]

I need to set the value of one column based on the value of another in a Pandas dataframe. This is the logic:
if df['c1'] == 'Value':
df['c2'] = 10
else:
df['c2'] = df['c3']
I am unable to get this to do what I want, which is to simply create a column with new values (or change the value of an existing column: either one works for me).
If I try to run the code above or if I write it as a function and use the apply method, I get the following:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

one way to do this would be to use indexing with .loc.
Example
In the absence of an example dataframe, I'll make one up here:
import numpy as np
import pandas as pd
df = pd.DataFrame({'c1': list('abcdefg')})
df.loc[5, 'c1'] = 'Value'
>>> df
c1
0 a
1 b
2 c
3 d
4 e
5 Value
6 g
Assuming you wanted to create a new column c2, equivalent to c1 except where c1 is Value, in which case, you would like to assign it to 10:
First, you could create a new column c2, and set it to equivalent as c1, using one of the following two lines (they essentially do the same thing):
df = df.assign(c2 = df['c1'])
# OR:
df['c2'] = df['c1']
Then, find all the indices where c1 is equal to 'Value' using .loc, and assign your desired value in c2 at those indices:
df.loc[df['c1'] == 'Value', 'c2'] = 10
And you end up with this:
>>> df
c1 c2
0 a a
1 b b
2 c c
3 d d
4 e e
5 Value 10
6 g g
If, as you suggested in your question, you would perhaps sometimes just want to replace the values in the column you already have, rather than create a new column, then just skip the column creation, and do the following:
df['c1'].loc[df['c1'] == 'Value'] = 10
# or:
df.loc[df['c1'] == 'Value', 'c1'] = 10
Giving you:
>>> df
c1
0 a
1 b
2 c
3 d
4 e
5 10
6 g

You can use np.where() to set values based on a specified condition:
#df
c1 c2 c3
0 4 2 1
1 8 7 9
2 1 5 8
3 3 3 5
4 3 6 8
Now change values (or set) in column ['c2'] based on your condition.
df['c2'] = np.where(df.c1 == 8,'X', df.c3)
c1 c2 c3
0 4 1 1
1 8 X 9
2 1 8 8
3 3 5 5
4 3 8 8

try:
df['c2'] = df['c1'].apply(lambda x: 10 if x == 'Value' else x)

Note the tilda that reverses the selection. It uses pandas methods (i.e. is faster than if/else).
df.loc[(df['c1'] == 'Value'), 'c2'] = 10
df.loc[~(df['c1'] == 'Value'), 'c2'] = df['c3']

I suggest doing it in two steps:
# set fixed value to 'c2' where the condition is met
df.loc[df['c1'] == 'Value', 'c2'] = 10
# copy value from 'c3' to 'c2' where the condition is NOT met
df.loc[df['c1'] != 'Value', 'c2'] = df[df['c1'] != 'Value', 'c3']

You can use pandas.DataFrame.mask to add virtually as many conditions as you need:
data = {'a': [1,2,3,4,5], 'b': [6,8,9,10,11]}
d = pd.DataFrame.from_dict(data, orient='columns')
c = {'c1': (2, 'Value1'), 'c2': (3, 'Value2'), 'c3': (5, d['b'])}
d['new'] = np.nan
for value in c.values():
d['new'].mask(d['a'] == value[0], value[1], inplace=True)
d['new'] = d['new'].fillna('Else')
d
Output:
a b new
0 1 6 Else
1 2 8 Value1
2 3 9 Value2
3 4 10 Else
4 5 11 11

I believe Series.map() to be very readable and efficient, e.g.:
df["c2"] = df["c1"].map(lambda x: 10 if x == 'Value' else x)
I like it because if the conditional logic gets more complex you can move it to a function and just pass in that function instead of the lambda.
If you need to base your conditional logic on more than one column you can use DataFrame.apply() as others suggest.

Try out df.apply() if you've a small/medium dataframe,
df['c2'] = df.apply(lambda x: 10 if x['c1'] == 'Value' else x['c1'], axis = 1)
Else, follow the slicing techniques mentioned in the above comments if you've got a big dataframe.

I had a big dataset and .loc[] was taking too long so I found a vectorized way to do it. Recall that you can set a column to a logical operator, so this works:
file['Flag'] = (file['Claim_Amount'] > 0)
This gives a Boolean, which I wanted, but you can multiply it by, say, 1 to make an Integer.

Element-wise Comparison of Two Pandas Dataframes

I am trying to compare two columns in pandas. I know I can do:
# either using Pandas' equals()
df1[col].equals(df2[col])
# or this
df1[col] == df2[col]
However, what I am looking for is to compare these columns elment-wise and when they are not matching print out both values. I have tried:
if df1[col] != df2[col]:
print(df1[col])
print(df2[col])
where I get the error for 'The truth value of a Series is ambiguous'
I believe this is because the column is treated as a series of boolean values for the comparison which causes the ambiguity. I also tried various forms of for loops which did not resolve the issue.
Can anyone point me to how I should go about doing what I described?

This might work for you:
import pandas as pd
df1 = pd.DataFrame({'col1': [1, 2, 3, 4, 5]})
df2 = pd.DataFrame({'col1': [1, 2, 9, 4, 7]})
if not df2[df2['col1'] != df1['col1']].empty:
print(df1[df1['col1'] != df2['col1']])
print(df2[df2['col1'] != df1['col1']])
Output:
col1
2 3
4 5
col1
2 9
4 7

You need to get hold of the index where the column values are not matching. Once you have that index then you can query the individual DFs to get the values.
Please try the fallowing and is if this helps:
for ind in (df1.loc[df1['col1'] != df2['col1']].index):
x = df1.loc[df1.index == ind, 'col1'].values[0]
y = df2.loc[df2.index == ind, 'col1'].values[0]
print(x, y )

Solution
Try this. You could use any of the following one-line solutions.
# Option-1
df.loc[df.apply(lambda row: row[col1] != row[col2], axis=1), [col1, col2]]
# Option-2
df.loc[df[col1]!=df[col2], [col1, col2]]
Logic:
Option-1: We use pandas.DataFrame.apply() to evaluate the target columns row by row and pass the returned indices to df.loc[indices, [col1, col2]] and that returns the required set of rows where col1 != col2.
Option-2: We get the indices with df[col1] != df[col2] and the rest of the logic is the same as Option-1.
Dummy Data
I made the dummy data such that for indices: 2,6,8 we will find column 'a' and 'c' to be different. Thus, we want only those rows returned by the solution.
import numpy as np
import pandas as pd
a = np.arange(10)
c = a.copy()
c[[2,6,8]] = [0,20,40]
df = pd.DataFrame({'a': a, 'b': a**2, 'c': c})
print(df)
Output:
a b c
0 0 0 0
1 1 1 1
2 2 4 0
3 3 9 3
4 4 16 4
5 5 25 5
6 6 36 20
7 7 49 7
8 8 64 40
9 9 81 9
Applying the solution to the dummy data
We see that the solution proposed returns the result as expected.
col1, col2 = 'a', 'c'
result = df.loc[df.apply(lambda row: row[col1] != row[col2], axis=1), [col1, col2]]
print(result)
Output:
a c
2 2 0
6 6 20
8 8 40

Find all duplicate columns in a collection of data frames

Having a collection of data frames, the goal is to identify the duplicated column names and return them as a list.
Example
The input are 3 data frames df1, df2 and df3:
df1 = pd.DataFrame({'a':[1,5], 'b':[3,9], 'e':[0,7]})
a b e
0 1 3 0
1 5 9 7
df2 = pd.DataFrame({'d':[2,3], 'e':[0,7], 'f':[2,1]})
d e f
0 2 0 2
1 3 7 1
df3 = pd.DataFrame({'b':[3,9], 'c':[8,2], 'e':[0,7]})
b c e
0 3 8 0
1 9 2 7
The output is a list [b, e]

pd.Series.duplicated
Since you are using Pandas, you can use pd.Series.duplicated after concatenating column names:
# concatenate column labels
s = pd.concat([df.columns.to_series() for df in (df1, df2, df3)])
# keep all duplicates only, then extract unique names
res = s[s.duplicated(keep=False)].unique()
print(res)
array(['b', 'e'], dtype=object)
pd.Series.value_counts
Alternatively, you can extract a series of counts and identify rows which have a count greater than 1:
s = pd.concat([df.columns.to_series() for df in (df1, df2, df3)]).value_counts()
res = s[s > 1].index
print(res)
Index(['e', 'b'], dtype='object')
collections.Counter
The classic Python solution is to use collections.Counter followed by a list comprehension. Recall that list(df) returns the columns in a dataframe, so we can use this map and itertools.chain to produce an iterable to feed Counter.
from itertools import chain
from collections import Counter
c = Counter(chain.from_iterable(map(list, (df1, df2, df3))))
res = [k for k, v in c.items() if v > 1]

here is my code for this problem, for comparing with only two data frames, with out concat them.
def getDuplicateColumns(df1, df2):
df_compare = pd.DataFrame({'df1':df1.columns.to_list()})
df_compare["df2"] = ""
# Iterate over all the columns in dataframe
for x in range(df1.shape[1]):
# Select column at xth index.
col = df1.iloc[:, x]
# Iterate over all the columns in DataFrame from (x+1)th index till end
duplicateColumnNames = []
for y in range(df2.shape[1]):
# Select column at yth index.
otherCol = df2.iloc[:, y]
# Check if two columns at x y index are equal
if col.equals(otherCol):
duplicateColumnNames.append(df2.columns.values[y])
df_compare.loc[df_compare["df1"]==df1.columns.values[x], "df2"] = str(duplicateColumnNames)
return df_compare

Find column names when row element meets a criteria Pandas

This is a basic question. I've got a square array with the rows and columns summed up. Eg:
df = pd.DataFrame([[0,0,1,0], [0,0,1,0], [1,0,0,0], [0,1,0,0]], index = ["a","b","c","d"], columns = ["a","b","c","d"])
df["sumRows"] = df.sum(axis = 1)
df.loc["sumCols"] = df.sum()
This returns:
In [100]: df
Out[100]:
a b c d sumRows
a 0 0 1 0 1
b 0 0 1 0 1
c 1 0 0 0 1
d 0 1 0 0 1
sumCols 1 1 2 0 4
I need to find the column labels for the sumCols rows which matches 0. At the moment I am doing this:
[df.loc["sumCols"] == 0].index
But this return a strange index type object. All I want is a list of values that match this criteria i.e: ['d'] in this case.

There is two ways (the index object can be converted to an interable like a list).
Do that with the columns:
columns = df.columns[df.sum()==0]
columns = list(columns)
Or you can rotate the Dataframe and treat columns as rows:
list(df.T[df.T.sumCols == 0].index)

You can use a lambda expression to filter series and if you want a list instead of index as result, you can call .tolist() on the index object:
(df.loc['sumCols'] == 0)[lambda x: x].index.tolist()
# ['d']
Or:
df.loc['sumCols'][lambda x: x == 0].index.tolist()
# ['d']
Without explicitly creating the sumCols and if you want to check which column has sum of zero, you can do:
df.sum()[lambda x: x == 0].index.tolist()
# ['d']
Check rows:
df.sum(axis = 1)[lambda x: x == 0].index.tolist()
# []
Note: The lambda expression approach is as fast as the vectorized method for subsetting, functional style and can be written easily in a one-liner if you prefer.

Heres a simple method using query after transposing
df.T.query('sumCols == 0').index.tolist()

Pandas: how to find and concatenate values

I'm trying to replace and add some values in pandas dataframe object. I have to following code
import pandas as pd
df = pd.DataFrame.from_items([('A', ["va-lue", "value-%", "value"]), ('B', [4, 5, 6])])
print df
df['A'] = df['A'].str.replace('%', '_0')
print df
df['A'] = df['A'].str.replace('-', '')
print df
#allmost there?
df.A[df['A'].str.contains('-')] + "_0"
How can I find the cell values in column A which contains '-' sign, replace this value with '' and add for these values a trailing '_0'? The resulting data set should look like this
A B
0 value_0 4
1 value_0 5
2 value 6

You can first keep track of the rows whose A needs to be appended with the trailing string, and perform these operations in two steps:
mask = df['A'].str.contains('-')
df['A'] = df['A'].str.replace('-|%', '')
df.ix[mask, 'A'] += '_0'
print df
Output:
A B
0 value_0 4
1 value_0 5
2 value 6