TypeError:不支持python 3.x Anacond中-：'str'和'str'的操作数类型

Question

我试图在一个大数据集中计算每小时的一些实例。下面的代码似乎在Python2.7上运行良好，但我不得不将其升级到3.x最新版本的python，并在Anaconda上更新了所有包。当我试图执行程序时，我得到了如下的str错误

代码：

import pandas as pd
from datetime import datetime,time
import numpy as np

fn = r'00_input.csv'
cols = ['UserId', 'UserMAC', 'HotspotID', 'StartTime', 'StopTime']
df = pd.read_csv(fn, header=None, names=cols)

df['m'] = df.StopTime + df.StartTime
df['d'] = df.StopTime - df.StartTime

# 'start' and 'end' for the reporting DF: `r`
# which will contain equal intervals (1 hour in this case)
start = pd.to_datetime(df.StartTime.min(), unit='s').date()
end = pd.to_datetime(df.StopTime.max(), unit='s').date() + pd.Timedelta(days=1)

# building reporting DF: `r`
freq = '1H'  # 1 Hour frequency
idx = pd.date_range(start, end, freq=freq)
r = pd.DataFrame(index=idx)
r['start'] = (r.index - pd.datetime(1970,1,1)).total_seconds().astype(np.int64)

# 1 hour in seconds, minus one second (so that we will not count it twice)
interval = 60*60 - 1

r['LogCount'] = 0
r['UniqueIDCount'] = 0

for i, row in r.iterrows():
        # intervals overlap test
        # https://en.wikipedia.org/wiki/Interval_tree#Overlap_test
        # i've slightly simplified the calculations of m and d
        # by getting rid of division by 2,
        # because it can be done eliminating common terms
    u = df[np.abs(df.m - 2*row.start - interval) < df.d + interval].UserID
    r.ix[i, ['LogCount', 'UniqueIDCount']] = [len(u), u.nunique()]

r['Date'] = pd.to_datetime(r.start, unit='s').dt.date
r['Day'] = pd.to_datetime(r.start, unit='s').dt.weekday_name.str[:3]
r['StartTime'] = pd.to_datetime(r.start, unit='s').dt.time
r['EndTime'] = pd.to_datetime(r.start + interval + 1, unit='s').dt.time

#r.to_csv('results.csv', index=False)
#print(r[r.LogCount > 0])
#print (r['StartTime'], r['EndTime'], r['Day'], r['LogCount'], r['UniqueIDCount'])

rout =  r[['Date', 'StartTime', 'EndTime', 'Day', 'LogCount', 'UniqueIDCount'] ]
#print rout
rout.to_csv('o_1_hour.csv', index=False, header=False

（第页）

在何处进行更改以获得无错误执行

错误：

File "C:\Program Files\Anaconda3\lib\site-packages\pandas\core\ops.py", line 686, in <lambda>
    lambda x: op(x, rvalues))

TypeError: unsupported operand type(s) for -: 'str' and 'str'

谢谢你的帮助，提前谢谢

Answer 1

df['d'] = df.StopTime - df.StartTime正在尝试从另一个字符串中减去一个字符串。我不知道你的数据是什么样子的，但很可能你想把StopTime和StartTime解析为日期。试试看

df = pd.read_csv(fn, header=None, names=cols, parse_dates=[3,4])

而不是df = pd.read_csv(fn, header=None, names=cols)。

Answer 2

我认为您需要将header=0更改为select first row to header，然后用list cols替换列名。

如果仍然存在问题，则需要^{}，因为StartTime和StopTime中的某些值是字符串，被解析为NaN，替换为0最后一个转换列为int：

cols = ['UserId', 'UserMAC', 'HotspotID', 'StartTime', 'StopTime']
df = pd.read_csv('canada_mini_unixtime.csv', header=0, names=cols)
#print (df)

df['StartTime'] = pd.to_numeric(df['StartTime'], errors='coerce').fillna(0).astype(int)
df['StopTime'] =  pd.to_numeric(df['StopTime'], errors='coerce').fillna(0).astype(int)

无变化：

df['m'] = df.StopTime + df.StartTime
df['d'] = df.StopTime - df.StartTime
start = pd.to_datetime(df.StartTime.min(), unit='s').date()
end = pd.to_datetime(df.StopTime.max(), unit='s').date() + pd.Timedelta(days=1)

freq = '1H'  # 1 Hour frequency
idx = pd.date_range(start, end, freq=freq)
r = pd.DataFrame(index=idx)
r['start'] = (r.index - pd.datetime(1970,1,1)).total_seconds().astype(np.int64)

# 1 hour in seconds, minus one second (so that we will not count it twice)
interval = 60*60 - 1

r['LogCount'] = 0
r['UniqueIDCount'] = 0

ix在上一版本的pandas中不推荐使用，因此请使用loc，并且列名在[]：

for i, row in r.iterrows():
        # intervals overlap test
        # https://en.wikipedia.org/wiki/Interval_tree#Overlap_test
        # i've slightly simplified the calculations of m and d
        # by getting rid of division by 2,
        # because it can be done eliminating common terms
    u = df.loc[np.abs(df.m - 2*row.start - interval) < df.d + interval, 'UserId']
    r.loc[i, ['LogCount', 'UniqueIDCount']] = [len(u), u.nunique()]

r['Date'] = pd.to_datetime(r.start, unit='s').dt.date
r['Day'] = pd.to_datetime(r.start, unit='s').dt.weekday_name.str[:3]
r['StartTime'] = pd.to_datetime(r.start, unit='s').dt.time
r['EndTime'] = pd.to_datetime(r.start + interval + 1, unit='s').dt.time

print (r)