Python 在iloc NULL上设置值
下面我有一个示例数据帧和函数。我创建了一个函数，该函数将获取“cell”的坐标，并将其放入一个元组，以及将其放入元组的原因。我想让这个函数也改变某一列的值Python 在iloc NULL上设置值,python,pandas,dataframe,Python,Pandas,Dataframe,下面我有一个示例数据帧和函数。我创建了一个函数，该函数将获取“cell”的坐标，并将其放入一个元组，以及将其放入元组的原因。我想让这个函数也改变某一列的值 import pandas as pd import numpy as np df1 = pd.DataFrame({'A' : [np.NaN,np.NaN,3,4,5,5,3,1,5,np.NaN], 'B' : [1,0,3,5,0,0,np.NaN,9,0,0],
import pandas as pd
import numpy as np
df1 = pd.DataFrame({'A' : [np.NaN,np.NaN,3,4,5,5,3,1,5,np.NaN],
'B' : [1,0,3,5,0,0,np.NaN,9,0,0],
'C' : [10,0,30,50,0,0,4,10,1,0],
'D' : [1,0,3,4,0,0,7,8,0,1],
'E' : [np.nan,'Unassign','Assign','Ugly','Appreciate',
'Undo','Assign','Unicycle','Assign','Unicorn',]})
print(df1)
highlights = []
def find_nan(list_col):
for c in list_col:
# if column is one of the dataframe's columns, go
if c in df1.columns:
# for each index x where column c of the dataframe is null, go
for x in df1.loc[df1[c].isnull()].index: #appends to the list of tuples
highlights.append(
tuple([x + 2, df1.columns.get_loc(c) + 1, f'{c} is Null in row {x + 2}']))
df1.iloc[x, df1.columns.get_loc('E')] = f'{c} is blank in row {x + 2}'
find_nan(['A','B'])
# using the function above, checks for all nulls in A and B
# Also places the coordinates and reason in a tuple and changes values of column 'E'
#output:
A B C D E
0 NaN 1.0 10 1 A is blank in row 2
1 NaN 0.0 0 0 A is blank in row 3
2 3.0 3.0 30 3 Assign
3 4.0 5.0 50 4 Ugly
4 5.0 0.0 0 0 Appreciate
5 5.0 0.0 0 0 Undo
6 3.0 NaN 4 7 Assign
7 1.0 9.0 10 8 Unicycle
8 5.0 0.0 1 0 Assign
9 NaN 0.0 0 1 A is blank in row 11
EEdf1.ilocdf1.iloc[0]['E'].isnull（）AttributeError:'float'对象没有属性'isnull'np.isnan（df1.iloc[0]['E']）TrueETypeErrorif df1.iloc[x]['E'] is null:
df1.iloc[x, df1.columns.get_loc('E')] = 'PREVIOUS_VALUE' + f'{c} is blank in row {x + 2}'
else:
df1.iloc[x, df1.columns.get_loc('E')] = f'{c} is blank in row {x + 2}
find_nan(['A','B'])
A B C D E
0 NaN 1.0 10 1 A is blank in row 2
1 NaN 0.0 0 0 Unassign and A is blank in row 3
2 3.0 3.0 30 3 Assign
3 4.0 5.0 50 4 Ugly
4 5.0 0.0 0 0 Appreciate
5 5.0 0.0 0 0 Undo
6 3.0 NaN 4 7 Assign and B is blank in row 8
7 1.0 9.0 10 8 Unicycle
8 5.0 0.0 1 0 Assign
9 NaN 0.0 0 1 Unicorn and A is blank in row 11
使用Python3.6。这是一个具有更多功能的更大项目的一部分，因此“添加原因”和向索引中添加2“没有明显原因”注意，这是使用Python2测试的，但我没有注意到任何可能的情况 防止它在Python3中工作
def find_nan(df, cols):
if isinstance(cols, (str, unicode)):
cols = [cols] # Turn a single column into an list.
nulls = df[cols].isnull() # Find all null values in requested columns.
df['E'] = df['E'].replace(np.nan, "") # Turn NaN values into an empty string.
for col in cols:
if col not in df:
continue
# If null value in the column an existing value in column `E`, add " and ".
df.loc[(nulls[col] & df['E'].str.len().astype(bool)), 'E'] += ' and '
# For null column values, add to column `E`: "[Column name] is blank in row ".
df.loc[nulls[col], 'E'] += '{} is blank in row '.format(col)
# For null column values, add to column `E` the index location + 2.
df.loc[nulls[col], 'E'] += (df['E'][nulls[col]].index + 2).astype(str)
return df
>>> find_nan(df1, ['A', 'B'])
A B C D E
0 NaN 1 10 1 A is blank in row 2
1 NaN 0 0 0 Unassign and A is blank in row 3
2 3 3 30 3 Assign
3 4 5 50 4 Ugly
4 5 0 0 0 Appreciate
5 5 0 0 0 Undo
6 3 NaN 4 7 Assign and B is blank in row 8
7 1 9 10 8 Unicycle
8 5 0 1 0 Assign
9 NaN 0 0 1 Unicorn and A is blank in row 11
避免循环的可能解决方案
def val(ser):
row_value = ser['index']
check = ser[['A','B']].isnull()
found = check[check == True]
if len(found) == 1:
found = found.index[0]
if pd.isnull(ser['E']) == True:
return found + ' is blank in row ' + str(row_value + 2)
else:
return str(ser['E']) + ' and ' + found +' is blank in row ' + str(row_value+2)
else:
return ser['E']
df1['index'] = df1.index
df1['E'] = df1.apply(lambda row: val(row),axis=1)
print(df1.iloc[:,:5])
A B C D E
0 NaN 1.0 10 1 A is blank in row 2
1 NaN 0.0 0 0 Unassign and A is blank in row 3
2 3.0 3.0 30 3 Assign
3 4.0 5.0 50 4 Ugly
4 5.0 0.0 0 0 Appreciate
5 5.0 0.0 0 0 Undo
6 3.0 NaN 4 7 Assign and B is blank in row 8
7 1.0 9.0 10 8 Unicycle
8 5.0 0.0 1 0 Assign
9 NaN 0.0 0 1 Unicorn and A is blank in row 11
def val(ser):
check = ser[['A','B']].isnull()
found = check[check].index
if len(found) == 0:
return str(ser['E'])
else:
one = (str(ser['E'])+' and ').replace('nan and','')
two = ' and '.join([str(x) for x in found])
three = ' is blank in row ' + str(ser['index']+2)
return (one+two+three).strip()
df1['index'] = df1.index
df1['E'] = df1.apply(lambda row: val(row),axis=1)
print(df1.iloc[:,:5])
A B C D E
0 NaN NaN NaN 1 A and B is blank in row 2
1 NaN 0.0 0.0 0 Unassign and A is blank in row 3
2 3.0 3.0 30.0 3 Assign
3 4.0 5.0 50.0 4 Ugly
4 5.0 0.0 0.0 0 Appreciate
5 5.0 0.0 0.0 0 Undo
6 3.0 NaN 4.0 7 Assign and B is blank in row 8
7 1.0 9.0 10.0 8 Unicycle
8 5.0 0.0 1.0 0 Assign
9 NaN 0.0 0.0 1 Unicorn and A is blank in row 11
避免了Tubleed。如果一行有多个
NaN==True检查[check]