在列表中值的不同列标题中打印值

Question

此代码用于跟踪哪些交货延迟。我希望找到每个延迟交货的实例并找到与之关联的采购订单号。我当前的代码能够生成一个列表，显示指定范围内迟到的天数。但是，由于这些值存储在列表中，我无法找到与后期相关的采购订单编号。我希望在终端打印采购订单号和交货延迟天数。 （从那里我将使用 if 语句来仅查看大于 0 的值，以仅查看哪些交货延迟）。我的问题是，我如何在终端中打印采购订单号以及它旁边的延迟天数（我不知道如何执行此操作，因为“延迟天数”的所有值都存储在列表）

import csv
import pandas as pd
import datetime


def calculate(number):   
    fileread = pd.read_csv('otd.csv', encoding='latin-1')
    Deliveryvalue = fileread['Delivery Date']
    Desiredvalue = fileread['source desired delivery date']



    date_format = '%m/%d/%Y'

    date1 = datetime.datetime.strptime(Deliveryvalue[number], date_format)
    date2 = datetime.datetime.strptime(Desiredvalue[number], date_format)

    diff= date1 - date2
    diff2 = diff.days



    return diff2

list = [] 
for i in range(1,20):
    list.append(calculate(i))

for y in list: 
    if y > 1:
        print(list)    

打印这个：

[0, 0, 0, 0, 0, 0, 0, 0, -7, 3, 50, 0, 0, 0, 0, 0, 1, -9, 0]
# the negative numbers are early deliveries 

这是我的 csv 文件的一个虚拟示例：

Answer 1

import pandas as pd

# change names appropriately
PURCHASE_ORDER = 'Purchase Order'
DELIVERY_DATE = 'Delivery Date'
DESIRED_DATE = 'Desired Date'
DELAYED_DAYS = 'Delayed Days'

df = pd.read_csv('otd.csv', index_col=PURCHASE_ORDER)

-

>> df
                  Delivery Date Desired Date
Purchase Order
001               2014-12-31   2014-12-31
002               2014-12-31   2014-12-31
003               2015-01-05   2015-01-05
004               2015-01-05   2015-01-05
005               2015-02-12   2015-02-11
006               2016-02-13   2016-02-11

最后两次交货迟到了。

df[DELIVERY_DATE] = pd.to_datetime(df[DELIVERY_DATE])
df[DESIRED_DATE] = pd.to_datetime(df[DESIRED_DATE])
df[DELAYED_DAYS] = df[DELIVERY_DATE] - df[DESIRED_DATE]
late_threshold = pd.Timedelta(days=0)
late_deliveries = df[DELAYED_DAYS] > late_threshold

-

>> df[late_deliveries].drop([DELIVERY_DATE, DESIRED_DATE], axis=1)

                     Delayed Days
Purchase Order             
005                  1 days
006                  2 days

Answer 2

似乎您希望它成为 calculate() 函数的一部分，以便您可以在其他东西上运行该函数。也许尝试做这样的事情：

data = {}
data['ordernum'] = 'ordernum'
data['delayed_days'] = 'diff2'

return data

因此，每次您在数据帧的一行上循环它时，它基本上都会返回一个 python 字典（json）。

这是我玩过的代码：

import csv
import pandas as pd
import datetime


def calculate(row):   
    Deliveryvalue = row['delivery']
    Desiredvalue = row['desired']

    date_format = '%m/%d/%Y'

    date1 = datetime.datetime.strptime(Deliveryvalue, date_format)
    date2 = datetime.datetime.strptime(Desiredvalue, date_format)

    diff= date1 - date2
    diff2 = diff.days

    data = {}
    data['ordernum'] = row['order']
    data['delayed_days'] = diff2

    return data

file = pd.read_csv('otd.csv')

l = []

for index, row in file.iterrows():
    data = calculate(row)
    l.append(data)

print(l)

我改变了函数的工作方式，所以现在它运行在数据帧的行中。如果我对您的问题的解释正确，这应该是解决方案。

输出：

{'ordernum': 1, 'delayed_days': 0}
{'ordernum': 2, 'delayed_days': 0}
{'ordernum': 3, 'delayed_days': 0}
{'ordernum': 4, 'delayed_days': 0}
{'ordernum': 5, 'delayed_days': 0}

Answer 3

我采用了不同的方法，将您的列拉上拉链，然后进行比较。 抱歉，标题看起来很不稳定，我看不到整件事。我在您的 csv 文件中添加了 2 行以包含延迟交货。这些命令是：

Purchase order number    Delivery Date    Source Desired Deliv
17                       2/10/2018        2/5/2018
18                       7/16/2017        7/14/2018

见下文：

import pandas as pd
from datetime import datetime
from datetime import timedelta
import csv

df = pd.read_csv('./Desktop/dummy.csv')
late_items = []
date_format = '%m/%d/%Y'

for x,y,z in zip(df['Purchase order number'], df['Delivery Date'], df['Source desired delive']):
    actual_deliv_date = datetime.strptime(y, date_format)       
    supposed_deliv_date = datetime.strptime(z, date_format)    
    diff_deliv_date = supposed_deliv_date - actual_deliv_date
    if diff_deliv_date < timedelta(0):
        late_items.append([x, diff_deliv_date]) 
print(late_items)

输出：

[[17, datetime.timedelta(-5)], [18, datetime.timedelta(-2)]]

或者这样做只是将“Diff Delivery Date”列添加到您的原始 df：

diff_delivery_date = []
date_format = '%m/%d/%Y'
for x,y,z in zip(df['Purchase order number'], df['Delivery Date'], df['Source desired delive']):
    actual_deliv_date = datetime.strptime(y, date_format)
    supposed_deliv_date = datetime.strptime(z, date_format)
    diff_deliv_date = supposed_deliv_date - actual_deliv_date
    diff_delivery_date.append(diff_deliv_date)

df['Diff Deliv Date'] = diff_delivery_date
df.loc[df['Diff Deliv Date'] < timedelta(0)] # To get only those values less than 0 for late deliveries.
#df option to output whole df with on time and late deliveries.

输出：

    Purchase order number Delivery Date Source desired delive Diff Deliv Date

5                     17     2/10/2018              2/5/2018         -5 days
6                     18     7/16/2017             7/14/2017         -2 days