I have a problem with my Python 3 program. I use Mac OS X. This code is running properly.
# -*- coding: utf-8 -*-
#! python3
# sendDuesReminders.py - Sends e
You probably want to assign the lastname
you are reading out here
lastname = sheet.cell(row=r, column=3).value
to something; currently the program just forgets it
you could do that two lines after, like so
unpaidMembers[name] = lastname, email
your program will still crash at the same place, because .items()
still won't give you 3-tuples but rather something that has this structure: (name, (lastname, email))
good news is, python can handle this
for name, (lastname, email) in unpaidMembers.items():
etc.
1. First should understand the error meaning
Error not enough values to unpack (expected 3, got 2)
means:
a 2 part tuple, but assign to 3 values
and I have written demo code to show for you:
#!/usr/bin/python
# -*- coding: utf-8 -*-
# Function: Showing how to understand ValueError 'not enough values to unpack (expected 3, got 2)'
# Author: Crifan Li
# Update: 20191212
def notEnoughUnpack():
"""Showing how to understand python error `not enough values to unpack (expected 3, got 2)`"""
# a dict, which single key's value is two part tuple
valueIsTwoPartTupleDict = {
"name1": ("lastname1", "email1"),
"name2": ("lastname2", "email2"),
}
# Test case 1: got value from key
gotLastname, gotEmail = valueIsTwoPartTupleDict["name1"] # OK
print("gotLastname=%s, gotEmail=%s" % (gotLastname, gotEmail))
# gotLastname, gotEmail, gotOtherSomeValue = valueIsTwoPartTupleDict["name1"] # -> ValueError not enough values to unpack (expected 3, got 2)
# Test case 2: got from dict.items()
for eachKey, eachValues in valueIsTwoPartTupleDict.items():
print("eachKey=%s, eachValues=%s" % (eachKey, eachValues))
# same as following:
# Background knowledge: each of dict.items() return (key, values)
# here above eachValues is a tuple of two parts
for eachKey, (eachValuePart1, eachValuePart2) in valueIsTwoPartTupleDict.items():
print("eachKey=%s, eachValuePart1=%s, eachValuePart2=%s" % (eachKey, eachValuePart1, eachValuePart2))
# but following:
for eachKey, (eachValuePart1, eachValuePart2, eachValuePart3) in valueIsTwoPartTupleDict.items(): # will -> ValueError not enough values to unpack (expected 3, got 2)
pass
if __name__ == "__main__":
notEnoughUnpack()
using VSCode
debug effect:
2. For your code
for name, email, lastname in unpaidMembers.items():
but error
ValueError: not enough values to unpack (expected 3, got 2)
means each item(a tuple value) in unpaidMembers
, only have 1 parts:email
, which corresponding above code
unpaidMembers[name] = email
so should change code to:
for name, email in unpaidMembers.items():
to avoid error.
But obviously you expect extra lastname
, so should change your above code to
unpaidMembers[name] = (email, lastname)
and better change to better syntax:
for name, (email, lastname) in unpaidMembers.items():
then everything is OK and clear.
In this line:
for name, email, lastname in unpaidMembers.items():
unpaidMembers.items()
must have only two values per iteration.
Here is a small example to illustrate the problem:
This will work:
for alpha, beta, delta in [("first", "second", "third")]:
print("alpha:", alpha, "beta:", beta, "delta:", delta)
This will fail, and is what your code does:
for alpha, beta, delta in [("first", "second")]:
print("alpha:", alpha, "beta:", beta, "delta:", delta)
In this last example, what value in the list is assigned to delta
? Nothing, There aren't enough values, and that is the problem.
Since unpaidMembers
is a dictionary it always returns two values when called with .items()
- (key, value). You may want to keep your data as a list of tuples [(name, email, lastname), (name, email, lastname)..]
.
ValueErrors :In Python, a value is the information that is stored within a certain object. To encounter a ValueError in Python means that is a problem with the content of the object you tried to assign the value to.
in your case name,lastname and email 3 parameters are there but unpaidmembers only contain 2 of them.
name, lastname, email in unpaidMembers.items() so you should refer data or your code might be
lastname, email in unpaidMembers.items() or name, email in unpaidMembers.items()